Split (1)

train · 78.8k rows

description stringlengths 171 4k	code stringlengths 94 3.98k	normalized_code stringlengths 57 4.99k
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys. The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank. There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline. What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0. -----Input----- The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys. -----Output----- Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1. -----Examples----- Input 6 9 2 4 Output 4 Input 6 10 2 4 Output 2 Input 6 5 4 3 Output -1 -----Note----- In the first example the bus needs to refuel during each journey. In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty. In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.	a, b, f, k = list(map(int, input().split(" "))) fuel_level = b point = 0 direction = 1 journays = 0 impossible = 0 refuel_counter = 0 if k == 1 and (b < a - f or b < f): print(-1) elif k == 2 and (b < f or b < 2 * (a - f)): print(-1) elif k > 2 and (b < 2 * f or b < 2 * (a - f)): print(-1) else: while journays < k: if point == 0 and direction == 1: fuel_level -= f point = f elif point == f and direction == 1: if ( fuel_level < 2 * (a - f) and journays < k - 1 or fuel_level < a - f and journays == k - 1 ): refuel_counter += 1 fuel_level = b - (a - f) journays += 1 direction = -1 point = a else: fuel_level -= a - f journays += 1 direction = -1 point = a elif point == a and direction == -1: fuel_level -= a - f point = f elif point == f and direction == -1: if ( fuel_level < 2 * f and journays < k - 1 or fuel_level < f and journays == k - 1 ): refuel_counter += 1 fuel_level = b - f journays += 1 direction = 1 point = 0 else: fuel_level -= f journays += 1 direction = 1 point = 0 print(refuel_counter)	ASSIGN VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR NUMBER VAR VAR VAR BIN_OP NUMBER BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR NUMBER VAR BIN_OP NUMBER VAR VAR BIN_OP NUMBER BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER WHILE VAR VAR IF VAR NUMBER VAR NUMBER VAR VAR ASSIGN VAR VAR IF VAR VAR VAR NUMBER IF VAR BIN_OP NUMBER BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR IF VAR VAR VAR NUMBER VAR BIN_OP VAR VAR ASSIGN VAR VAR IF VAR VAR VAR NUMBER IF VAR BIN_OP NUMBER VAR VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER VAR VAR VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys. The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank. There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline. What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0. -----Input----- The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys. -----Output----- Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1. -----Examples----- Input 6 9 2 4 Output 4 Input 6 10 2 4 Output 2 Input 6 5 4 3 Output -1 -----Note----- In the first example the bus needs to refuel during each journey. In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty. In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.	a, b, f, k = map(int, input().split()) d = b n = 0 s = 0 x = a - f d -= f if x * 2 > b or f * 2 > b: if k < 3: if k == 1: if a <= b: print(0) elif f <= b and x <= b: print(1) else: print(-1) if k == 2: if a + x <= b: print(1) elif f <= b and x * 2 <= b: print(2) else: print(-1) else: print(-1) else: while n < k: if n + 1 == k: if d < x: s += 1 break if d < x * 2: s += 1 d = b n += 1 d -= x * 2 if n + 1 == k: if d < f: s += 1 break if d < f * 2: s += 1 d = b n += 1 d -= f * 2 print(s)	ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR VAR IF BIN_OP VAR NUMBER VAR BIN_OP VAR NUMBER VAR IF VAR NUMBER IF VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR VAR VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF VAR NUMBER IF BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER WHILE VAR VAR IF BIN_OP VAR NUMBER VAR IF VAR VAR VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR NUMBER VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER VAR IF VAR VAR VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR NUMBER VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR VAR
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys. The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank. There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline. What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0. -----Input----- The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys. -----Output----- Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1. -----Examples----- Input 6 9 2 4 Output 4 Input 6 10 2 4 Output 2 Input 6 5 4 3 Output -1 -----Note----- In the first example the bus needs to refuel during each journey. In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty. In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.	def can_make_route(): can_return_to_station = ( bus["gasoline"] - (abs(bus["position"] - bus["direction"]) + abs(bus["direction"] - station)) >= 0 ) is_last_route = ( bus["qty_routs"] + 1 == cnt_routes and bus["gasoline"] - abs(bus["position"] - bus["direction"]) >= 0 ) return can_return_to_station or is_last_route def make_route(): bus["gasoline"] = bus["gasoline"] - abs(bus["position"] - bus["direction"]) bus["position"] = bus["direction"] bus["direction"] = start if bus["direction"] == end else end bus["qty_routs"] += 1 def goto_station(): bus["gasoline"] = max_gasoline bus["position"] = station def is_input_correct(): return max_gasoline - station >= 0 end, max_gasoline, station, cnt_routes = map(int, input().split()) start, res = 0, 0 bus = {"position": start, "gasoline": max_gasoline, "direction": end, "qty_routs": 0} if is_input_correct(): while bus["qty_routs"] != cnt_routes: if can_make_route(): make_route() elif bus["position"] == station: res = -1 break else: goto_station() res += 1 else: res = -1 print(res)	FUNC_DEF ASSIGN VAR BIN_OP VAR STRING BIN_OP FUNC_CALL VAR BIN_OP VAR STRING VAR STRING FUNC_CALL VAR BIN_OP VAR STRING VAR NUMBER ASSIGN VAR BIN_OP VAR STRING NUMBER VAR BIN_OP VAR STRING FUNC_CALL VAR BIN_OP VAR STRING VAR STRING NUMBER RETURN VAR VAR FUNC_DEF ASSIGN VAR STRING BIN_OP VAR STRING FUNC_CALL VAR BIN_OP VAR STRING VAR STRING ASSIGN VAR STRING VAR STRING ASSIGN VAR STRING VAR STRING VAR VAR VAR VAR STRING NUMBER FUNC_DEF ASSIGN VAR STRING VAR ASSIGN VAR STRING VAR FUNC_DEF RETURN BIN_OP VAR VAR NUMBER ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR NUMBER NUMBER ASSIGN VAR DICT STRING STRING STRING STRING VAR VAR VAR NUMBER IF FUNC_CALL VAR WHILE VAR STRING VAR IF FUNC_CALL VAR EXPR FUNC_CALL VAR IF VAR STRING VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys. The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank. There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline. What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0. -----Input----- The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys. -----Output----- Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1. -----Examples----- Input 6 9 2 4 Output 4 Input 6 10 2 4 Output 2 Input 6 5 4 3 Output -1 -----Note----- In the first example the bus needs to refuel during each journey. In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty. In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.	a, b, f, k = map(int, input().split(" ")) if ( k > 2 and (b < 2 * f or b < 2 * (a - f)) or k == 1 and (b < f or b < a - f) or k == 2 and (b < f or b < 2 * (a - f)) ): print(-1) else: state = 2 cir = 0 rb = b - f c = 0 while cir < k: if state == 1: if cir + 1 == k: if rb < f: c += 1 break else: if rb < 2 * f: rb = b c += 1 rb -= 2 * f cir += 1 state = 2 elif state == 2: if cir + 1 == k: if rb < a - f: c += 1 break else: if rb < 2 * (a - f): rb = b c += 1 rb -= 2 * (a - f) state = 1 cir += 1 print(c)	ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING IF VAR NUMBER VAR BIN_OP NUMBER VAR VAR BIN_OP NUMBER BIN_OP VAR VAR VAR NUMBER VAR VAR VAR BIN_OP VAR VAR VAR NUMBER VAR VAR VAR BIN_OP NUMBER BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR NUMBER WHILE VAR VAR IF VAR NUMBER IF BIN_OP VAR NUMBER VAR IF VAR VAR VAR NUMBER IF VAR BIN_OP NUMBER VAR ASSIGN VAR VAR VAR NUMBER VAR BIN_OP NUMBER VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER IF BIN_OP VAR NUMBER VAR IF VAR BIN_OP VAR VAR VAR NUMBER IF VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR VAR VAR NUMBER VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys. The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank. There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline. What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0. -----Input----- The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys. -----Output----- Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1. -----Examples----- Input 6 9 2 4 Output 4 Input 6 10 2 4 Output 2 Input 6 5 4 3 Output -1 -----Note----- In the first example the bus needs to refuel during each journey. In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty. In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.	a = 0 b = 0 f = 0 k = 0 def get_user_input(): global a global b global f global k interval_info = input("").split() a = int(interval_info[0]) b = int(interval_info[1]) f = int(interval_info[2]) k = int(interval_info[3]) def calc(): global b m = 0 count = 0 B = b if B < a - f or B < f: return -1 while m + 1 < k: if m % 2 == 0 and b < a + (a - f): count += 1 m += 1 b = B - (a - f) if b < a - f: return -1 elif m % 2 != 0 and b < a + f: count += 1 m += 1 b = B - f if b < f: return -1 else: m += 1 b = b - a if b - a < 0: count += 1 return count get_user_input() print(calc())	ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FUNC_DEF ASSIGN VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER FUNC_DEF ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR IF VAR BIN_OP VAR VAR VAR VAR RETURN NUMBER WHILE BIN_OP VAR NUMBER VAR IF BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR BIN_OP VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR IF VAR BIN_OP VAR VAR RETURN NUMBER IF BIN_OP VAR NUMBER NUMBER VAR BIN_OP VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF VAR VAR RETURN NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF BIN_OP VAR VAR NUMBER VAR NUMBER RETURN VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys. The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank. There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline. What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0. -----Input----- The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys. -----Output----- Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1. -----Examples----- Input 6 9 2 4 Output 4 Input 6 10 2 4 Output 2 Input 6 5 4 3 Output -1 -----Note----- In the first example the bus needs to refuel during each journey. In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty. In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.	inp = input().split(" ") a = int(inp[0]) b = int(inp[1]) f = int(inp[2]) k = int(inp[3]) b_temp = b i = 0 doz = 0 if k > 2 and b < 2 * f or k > 1 and b < 2 * (a - f): print(-1) else: while i < k: if k == 1 and max(f, a - f) > b: print(-1) break else: b = b - f if k - i == 1: if b < a - f: doz = doz + 1 i = i + 1 b = b_temp b = b - (a - f) elif b == a - f: i = i + 1 b = b - (a - f) elif b < 2 * (a - f) and b >= a - f: i = i + 1 b = b_temp b = b - (a - f) elif b >= 2 * (a - f): b = b - (a - f) i = i + 1 elif b < 2 * (a - f): doz = doz + 1 i = i + 1 b = b_temp b = b - (a - f) elif b >= 2 * (a - f): b = b - (a - f) i = i + 1 if i == k: print(doz) break b = b - (a - f) if k - i == 1: if b < f: doz = doz + 1 i = i + 1 b = b_temp b = b - f elif b == f: i = i + 1 b = b - f elif b < 2 * f and b >= f: i = i + 1 b = b_temp b = b - f elif b >= 2 * f: b = b - f i = i + 1 elif b < 2 * f: doz = doz + 1 i = i + 1 b = b_temp b = b - f elif b >= 2 * f: b = b - f i = i + 1 if i == k: print(doz)	ASSIGN VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER VAR BIN_OP NUMBER VAR VAR NUMBER VAR BIN_OP NUMBER BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER WHILE VAR VAR IF VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF BIN_OP VAR VAR NUMBER IF VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR IF VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR IF VAR BIN_OP NUMBER BIN_OP VAR VAR VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR IF VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR IF VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR IF BIN_OP VAR VAR NUMBER IF VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF VAR BIN_OP NUMBER VAR VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR BIN_OP NUMBER VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR BIN_OP NUMBER VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR BIN_OP NUMBER VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR VAR
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys. The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank. There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline. What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0. -----Input----- The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys. -----Output----- Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1. -----Examples----- Input 6 9 2 4 Output 4 Input 6 10 2 4 Output 2 Input 6 5 4 3 Output -1 -----Note----- In the first example the bus needs to refuel during each journey. In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty. In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.	a, b, f, k = map(int, input().split()) tf = a - f gas = b - f if gas < 0: print(-1) exit() ans = 0 for i in range(k): if i % 2 == 0: if i == k - 1: if gas < tf: gas = b ans += 1 elif gas < tf * 2: gas = b ans += 1 if i == k - 1: gas -= tf else: gas -= tf * 2 if gas < 0: print(-1) exit() else: if i == k - 1: if gas < f: gas = b ans += 1 elif gas < f * 2: gas = b ans += 1 if i == k - 1: gas -= f else: gas -= f * 2 if gas < 0: print(-1) exit() print(ans)	ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER IF VAR BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR VAR VAR NUMBER IF VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR IF VAR BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR VAR VAR NUMBER IF VAR BIN_OP VAR NUMBER ASSIGN VAR VAR VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys. The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank. There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline. What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0. -----Input----- The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys. -----Output----- Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1. -----Examples----- Input 6 9 2 4 Output 4 Input 6 10 2 4 Output 2 Input 6 5 4 3 Output -1 -----Note----- In the first example the bus needs to refuel during each journey. In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty. In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.	def solve(): a, b, f, k = [int(st) for st in input().split(" ")] fuel = b location = 0 direction = "right" fueled = 0 journeypassed = 0 while journeypassed < k: if location == 0: if fuel < f: return -1 fuel -= f location = f direction = "right" elif location == a: if fuel < a - f: return -1 fuel -= a - f location = f direction = "left" elif location == f: if k - journeypassed <= 1: if direction == "left": if fuel < f: fuel = b fueled += 1 if fuel < f: return -1 fuel -= f location = 0 direction = "right" else: if fuel < a - f: fuel = b fueled += 1 if fuel < a - f: return -1 fuel -= a - f location = a direction = "left" elif direction == "left": if fuel < 2 * f: fuel = b fueled += 1 if fuel < 2 * f: return -1 fuel -= 2 * f direction = "right" else: if fuel < 2 * (a - f): fuel = b fueled += 1 if fuel < 2 * (a - f): return -1 fuel -= 2 * (a - f) direction = "left" journeypassed += 1 return fueled print(solve())	FUNC_DEF ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR IF VAR NUMBER IF VAR VAR RETURN NUMBER VAR VAR ASSIGN VAR VAR ASSIGN VAR STRING IF VAR VAR IF VAR BIN_OP VAR VAR RETURN NUMBER VAR BIN_OP VAR VAR ASSIGN VAR VAR ASSIGN VAR STRING IF VAR VAR IF BIN_OP VAR VAR NUMBER IF VAR STRING IF VAR VAR ASSIGN VAR VAR VAR NUMBER IF VAR VAR RETURN NUMBER VAR VAR ASSIGN VAR NUMBER ASSIGN VAR STRING IF VAR BIN_OP VAR VAR ASSIGN VAR VAR VAR NUMBER IF VAR BIN_OP VAR VAR RETURN NUMBER VAR BIN_OP VAR VAR ASSIGN VAR VAR ASSIGN VAR STRING IF VAR STRING IF VAR BIN_OP NUMBER VAR ASSIGN VAR VAR VAR NUMBER IF VAR BIN_OP NUMBER VAR RETURN NUMBER VAR BIN_OP NUMBER VAR ASSIGN VAR STRING IF VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR VAR VAR NUMBER IF VAR BIN_OP NUMBER BIN_OP VAR VAR RETURN NUMBER VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR STRING VAR NUMBER RETURN VAR EXPR FUNC_CALL VAR FUNC_CALL VAR
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys. The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank. There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline. What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0. -----Input----- The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys. -----Output----- Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1. -----Examples----- Input 6 9 2 4 Output 4 Input 6 10 2 4 Output 2 Input 6 5 4 3 Output -1 -----Note----- In the first example the bus needs to refuel during each journey. In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty. In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.	b, max_fuel, f, k = map(int, input().split()) now_fuel = max_fuel ans = 0 part_1 = f part_2 = b - f now_fuel -= part_1 if now_fuel < 0: print("-1") exit() while k != 0: if k > 1: if now_fuel < part_2 * 2: ans += 1 now_fuel = max_fuel now_fuel -= part_2 * 2 k -= 1 if now_fuel < 0: print("-1") exit() else: if now_fuel < part_2: ans += 1 now_fuel = max_fuel now_fuel -= part_2 k -= 1 if now_fuel < 0: print("-1") exit() break if k > 1: if now_fuel < part_1 * 2: ans += 1 now_fuel = max_fuel now_fuel -= part_1 * 2 k -= 1 if now_fuel < 0: print("-1") exit() else: if now_fuel < part_1: ans += 1 now_fuel = max_fuel now_fuel -= part_1 k -= 1 if now_fuel < 0: print("-1") exit() break print(ans)	ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR WHILE VAR NUMBER IF VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR IF VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR IF VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR NUMBER VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR IF VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys. The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank. There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline. What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0. -----Input----- The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys. -----Output----- Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1. -----Examples----- Input 6 9 2 4 Output 4 Input 6 10 2 4 Output 2 Input 6 5 4 3 Output -1 -----Note----- In the first example the bus needs to refuel during each journey. In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty. In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.	a, b, f, k = (int(x) for x in input().split()) z = 0 curr_petrol = b curr_races_made = 0 ans = 0 z___f = f - z f___a = a - f if k == 1: if z___f > b or f___a > b: print(-1) return elif k == 2: if 2 * f___a > b: print(-1) return elif 2 * z___f > b or 2 * f___a > b: print(-1) return curr_save = None route = [z___f, f___a, f___a, z___f] while curr_races_made < 2 * k: curr_pos = curr_races_made % 4 curr_petrol -= route[curr_pos] curr_races_made += 1 if curr_petrol < 0: curr_petrol = b ans += 1 curr_races_made = curr_save continue if curr_pos == 0 or curr_pos == 2: curr_save = curr_races_made print(ans)	ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER IF VAR VAR VAR VAR EXPR FUNC_CALL VAR NUMBER RETURN IF VAR NUMBER IF BIN_OP NUMBER VAR VAR EXPR FUNC_CALL VAR NUMBER RETURN IF BIN_OP NUMBER VAR VAR BIN_OP NUMBER VAR VAR EXPR FUNC_CALL VAR NUMBER RETURN ASSIGN VAR NONE ASSIGN VAR LIST VAR VAR VAR VAR WHILE VAR BIN_OP NUMBER VAR ASSIGN VAR BIN_OP VAR NUMBER VAR VAR VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR VAR VAR NUMBER ASSIGN VAR VAR IF VAR NUMBER VAR NUMBER ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys. The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank. There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline. What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0. -----Input----- The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys. -----Output----- Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1. -----Examples----- Input 6 9 2 4 Output 4 Input 6 10 2 4 Output 2 Input 6 5 4 3 Output -1 -----Note----- In the first example the bus needs to refuel during each journey. In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty. In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.	a, b, f, k = list(map(int, input().split())) g = b ans = 0 p1, p2 = 0, f for i in range(2 * k): if p1 == 0 and p2 == f: if b >= p2 - p1: b = b - (p2 - p1) else: ans = -1 break p1 = f p2 = a elif p1 == f and p2 == a: y = 1 if i == 2 * k - 1 else 2 if b >= y * (p2 - p1): b = b - (p2 - p1) else: b = g b = b - (p2 - p1) if b < 0: ans = -1 break else: ans = ans + 1 p1 = a p2 = f elif p1 == a and p2 == f: if b >= p1 - p2: b = b - (p1 - p2) else: ans = -1 break p1 = f p2 = 0 elif p1 == f and p2 == 0: y = 1 if i == 2 * k - 1 else 2 if b >= y * (p1 - p2): b = b - (p1 - p2) else: b = g b = b - (p1 - p2) if b < 0: ans = -1 break else: ans = ans + 1 p1 = 0 p2 = f print(ans)	ASSIGN VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER VAR FOR VAR FUNC_CALL VAR BIN_OP NUMBER VAR IF VAR NUMBER VAR VAR IF VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR IF VAR VAR VAR VAR ASSIGN VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER NUMBER NUMBER IF VAR BIN_OP VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR IF VAR VAR VAR VAR IF VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER IF VAR VAR VAR NUMBER ASSIGN VAR VAR BIN_OP BIN_OP NUMBER VAR NUMBER NUMBER NUMBER IF VAR BIN_OP VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys. The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank. There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline. What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0. -----Input----- The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys. -----Output----- Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1. -----Examples----- Input 6 9 2 4 Output 4 Input 6 10 2 4 Output 2 Input 6 5 4 3 Output -1 -----Note----- In the first example the bus needs to refuel during each journey. In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty. In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.	def find_refuels(a, b, f, k): if b < f or b < a - f: return -1 if b < 2 * (a - f) and k > 1: return -1 if b < 2 * f and k > 2: return -1 curr_fuel = b - f refuel_count = 0 remaining_trips = k while remaining_trips: remaining_trips -= 1 if ( curr_fuel < 2 * (a - f) and remaining_trips or curr_fuel < a - f and not remaining_trips ): refuel_count += 1 curr_fuel = b curr_fuel -= 2 * (a - f) if not remaining_trips: break remaining_trips -= 1 if ( curr_fuel < 2 * f and remaining_trips or curr_fuel < f and not remaining_trips ): refuel_count += 1 curr_fuel = b curr_fuel -= 2 * f return refuel_count a, b, f, k = [int(x) for x in input().strip().split()] print(find_refuels(a, b, f, k))	FUNC_DEF IF VAR VAR VAR BIN_OP VAR VAR RETURN NUMBER IF VAR BIN_OP NUMBER BIN_OP VAR VAR VAR NUMBER RETURN NUMBER IF VAR BIN_OP NUMBER VAR VAR NUMBER RETURN NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR WHILE VAR VAR NUMBER IF VAR BIN_OP NUMBER BIN_OP VAR VAR VAR VAR BIN_OP VAR VAR VAR VAR NUMBER ASSIGN VAR VAR VAR BIN_OP NUMBER BIN_OP VAR VAR IF VAR VAR NUMBER IF VAR BIN_OP NUMBER VAR VAR VAR VAR VAR VAR NUMBER ASSIGN VAR VAR VAR BIN_OP NUMBER VAR RETURN VAR ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR VAR VAR VAR VAR
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys. The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank. There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline. What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0. -----Input----- The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys. -----Output----- Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1. -----Examples----- Input 6 9 2 4 Output 4 Input 6 10 2 4 Output 2 Input 6 5 4 3 Output -1 -----Note----- In the first example the bus needs to refuel during each journey. In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty. In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.	def main(): A, B, F, K = map(int, input().split()) gas = B refuel = 0 i = 1 while i <= K: if i % 2: gas -= F if gas >= 0: if K - i > 0 and gas < (A - F) * 2 or K - i == 0 and gas < A - F: gas = B refuel += 1 gas -= A - F else: gas -= A - F if gas >= 0: if K - i > 0 and gas < F * 2 or K - i == 0 and gas < F: gas = B refuel += 1 gas -= F if gas < 0: print(-1) return i += 1 print(refuel) main()	FUNC_DEF ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR IF BIN_OP VAR NUMBER VAR VAR IF VAR NUMBER IF BIN_OP VAR VAR NUMBER VAR BIN_OP BIN_OP VAR VAR NUMBER BIN_OP VAR VAR NUMBER VAR BIN_OP VAR VAR ASSIGN VAR VAR VAR NUMBER VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR IF VAR NUMBER IF BIN_OP VAR VAR NUMBER VAR BIN_OP VAR NUMBER BIN_OP VAR VAR NUMBER VAR VAR ASSIGN VAR VAR VAR NUMBER VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER RETURN VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys. The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank. There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline. What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0. -----Input----- The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys. -----Output----- Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1. -----Examples----- Input 6 9 2 4 Output 4 Input 6 10 2 4 Output 2 Input 6 5 4 3 Output -1 -----Note----- In the first example the bus needs to refuel during each journey. In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty. In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.	a, b, f, k = input().split() a, b, f, k = int(a), int(b), int(f), int(k) if k > 2 and b < max(2 * f, 2 * (a - f)): print(-1) exit() if k == 1 and b < max(f, a - f): print(-1) exit() if k == 2 and (b < 2 * (a - f) or b < f): print(-1) exit() v = b count = 0 while k > 0: v = v - f if v < 0: count = -1 break if k > 1 and v < 2 * (a - f): v = b count += 1 elif k == 1 and v < a - f: v = b count += 1 if k > 1: v = v - 2 * (a - f) else: v = v - (a - f) k -= 1 if k > 1 and v < 2 * f: v = b count += 1 elif k == 1 and v < f: v = b count += 1 v = v - f k -= 1 print(count)	ASSIGN VAR VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER VAR FUNC_CALL VAR BIN_OP NUMBER VAR BIN_OP NUMBER BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR IF VAR NUMBER VAR FUNC_CALL VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR IF VAR NUMBER VAR BIN_OP NUMBER BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR NUMBER WHILE VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER IF VAR NUMBER VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR VAR VAR NUMBER IF VAR NUMBER VAR BIN_OP VAR VAR ASSIGN VAR VAR VAR NUMBER IF VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR VAR NUMBER IF VAR NUMBER VAR BIN_OP NUMBER VAR ASSIGN VAR VAR VAR NUMBER IF VAR NUMBER VAR VAR ASSIGN VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys. The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank. There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline. What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0. -----Input----- The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys. -----Output----- Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1. -----Examples----- Input 6 9 2 4 Output 4 Input 6 10 2 4 Output 2 Input 6 5 4 3 Output -1 -----Note----- In the first example the bus needs to refuel during each journey. In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty. In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.	a, b, f, k = map(int, input().split(" ")) l, r = f, a - f if b < l or b < r: print(-1) elif k > 1 and b < 2 * r: print(-1) elif k > 2 and b < 2 * l: print(-1) else: remain = b cnt = 0 i = 0 while i < k: remain -= l if i < k - 1 and remain < 2 * r: remain = b cnt += 1 elif i == k - 1 and remain < r: remain = b cnt += 1 remain -= 2 * r i += 1 if i == k: break if i < k - 1 and remain < 2 * l: remain = b cnt += 1 elif i == k - 1 and remain < l: remain = b cnt += 1 remain -= l i += 1 print(cnt)	ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR VAR VAR BIN_OP VAR VAR IF VAR VAR VAR VAR EXPR FUNC_CALL VAR NUMBER IF VAR NUMBER VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR NUMBER IF VAR NUMBER VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR VAR VAR IF VAR BIN_OP VAR NUMBER VAR BIN_OP NUMBER VAR ASSIGN VAR VAR VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR VAR VAR NUMBER VAR BIN_OP NUMBER VAR VAR NUMBER IF VAR VAR IF VAR BIN_OP VAR NUMBER VAR BIN_OP NUMBER VAR ASSIGN VAR VAR VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR ASSIGN VAR VAR VAR NUMBER VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys. The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank. There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline. What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0. -----Input----- The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys. -----Output----- Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1. -----Examples----- Input 6 9 2 4 Output 4 Input 6 10 2 4 Output 2 Input 6 5 4 3 Output -1 -----Note----- In the first example the bus needs to refuel during each journey. In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty. In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.	a, b, f, k = map(int, input().split()) c, v = b, 0 def t(d): global c, v if c < d: c, v = b, v + 1 if c < d: print(-1) exit() c -= d t(f) for i in range(k - 1): t(2 * f if i % 2 else 2 * (a - f)) t(a - f if k % 2 else f) print(v)	ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR NUMBER FUNC_DEF IF VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP NUMBER VAR BIN_OP NUMBER BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys. The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank. There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline. What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0. -----Input----- The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys. -----Output----- Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1. -----Examples----- Input 6 9 2 4 Output 4 Input 6 10 2 4 Output 2 Input 6 5 4 3 Output -1 -----Note----- In the first example the bus needs to refuel during each journey. In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty. In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.	a, b, f, k = map(int, input().split()) tank = b dist = [f, a - f] direction = 0 total = 0 res = 0 journeys = 0 while journeys < k: if total + tank >= k * a: break if tank < dist[direction]: res = -1 break tank -= dist[direction] total += dist[direction] if tank < 2 * dist[(direction + 1) % 2]: tank = b res += 1 tank -= dist[(direction + 1) % 2] total += dist[(direction + 1) % 2] direction = (direction + 1) % 2 journeys += 1 if tank < 0 or tank < dist[direction] and journeys < k: res = -1 break print(res)	ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR LIST VAR BIN_OP VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR IF BIN_OP VAR VAR BIN_OP VAR VAR IF VAR VAR VAR ASSIGN VAR NUMBER VAR VAR VAR VAR VAR VAR IF VAR BIN_OP NUMBER VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR VAR VAR NUMBER VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER VAR NUMBER IF VAR NUMBER VAR VAR VAR VAR VAR ASSIGN VAR NUMBER EXPR FUNC_CALL VAR VAR
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys. The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank. There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline. What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0. -----Input----- The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys. -----Output----- Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1. -----Examples----- Input 6 9 2 4 Output 4 Input 6 10 2 4 Output 2 Input 6 5 4 3 Output -1 -----Note----- In the first example the bus needs to refuel during each journey. In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty. In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.	a, b, f, k = map(int, input().split()) s = a - f tch = f counter = 0 c = 0 full = b flag = True while flag: for i in range(k): b -= tch if b < 0: c = -1 if b < 2 * s and i != k - 1: counter += 1 b = full elif i == k - 1: if b < s: counter += 1 b = full b -= s if b < 0: c = -1 flag = False tch, s = s, tch flag = False if c != -1: print(counter) else: print(-1)	ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER WHILE VAR FOR VAR FUNC_CALL VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER IF VAR BIN_OP NUMBER VAR VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR VAR IF VAR BIN_OP VAR NUMBER IF VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys. The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank. There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline. What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0. -----Input----- The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys. -----Output----- Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1. -----Examples----- Input 6 9 2 4 Output 4 Input 6 10 2 4 Output 2 Input 6 5 4 3 Output -1 -----Note----- In the first example the bus needs to refuel during each journey. In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty. In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.	a, b, f, k = map(int, input().split()) bv = 0 j = 0 fu = 0 x = 0 ch = True bv += b def a_fs(): if b >= f: return True return False def a_ej(x, j, b): if j % 2 == 0 and a - x <= b or j % 2 == 1 and x <= b: return True return False def a_nfs(x, j, b): if j % 2 == 0 and b >= 2 * (a - x) or j % 2 == 1 and b >= 2 * x: return True return False def a_ejbv(x, j, bv): if j % 2 == 0 and a - x <= bv or j % 2 == 1 and x <= bv: return True return False def a_nfsbv(x, j, bv): if j % 2 == 0 and bv >= 2 * (a - x) or j % 2 == 1 and bv >= 2 * x: return True return False def x_nfs(x, j): if j % 2 == 0: return 2 * (a - x) return 2 * x def x_ej(x, j): if j % 2 == 0: return a - x return x if a_fs(): x = f bv -= f if k > 2: if a_nfs(x, 0, b) and a_nfs(x, 1, b): while k - j > 1: if a_nfsbv(x, j, bv): bv -= x_nfs(x, j) j += 1 else: fu += 1 bv = b - x_nfs(x, j) j += 1 if a_ejbv(x, j, bv): print(fu) else: print(fu + 1) else: print("-1") elif k == 2: if a_nfs(x, j, b): if a_nfsbv(x, j, bv): j += 1 if a_ejbv(x, j, bv): print(fu) else: print(fu + 1) else: j += 1 fu += 1 bv = b - 2 * (a - f) if a_ejbv(x, j, bv): print(fu) else: print(fu + 1) else: print(-1) elif a_ej(x, j, b): if a_ejbv(x, j, bv): print(0) else: print(1) else: print("-1") else: print("-1")	ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER VAR VAR FUNC_DEF IF VAR VAR RETURN NUMBER RETURN NUMBER FUNC_DEF IF BIN_OP VAR NUMBER NUMBER BIN_OP VAR VAR VAR BIN_OP VAR NUMBER NUMBER VAR VAR RETURN NUMBER RETURN NUMBER FUNC_DEF IF BIN_OP VAR NUMBER NUMBER VAR BIN_OP NUMBER BIN_OP VAR VAR BIN_OP VAR NUMBER NUMBER VAR BIN_OP NUMBER VAR RETURN NUMBER RETURN NUMBER FUNC_DEF IF BIN_OP VAR NUMBER NUMBER BIN_OP VAR VAR VAR BIN_OP VAR NUMBER NUMBER VAR VAR RETURN NUMBER RETURN NUMBER FUNC_DEF IF BIN_OP VAR NUMBER NUMBER VAR BIN_OP NUMBER BIN_OP VAR VAR BIN_OP VAR NUMBER NUMBER VAR BIN_OP NUMBER VAR RETURN NUMBER RETURN NUMBER FUNC_DEF IF BIN_OP VAR NUMBER NUMBER RETURN BIN_OP NUMBER BIN_OP VAR VAR RETURN BIN_OP NUMBER VAR FUNC_DEF IF BIN_OP VAR NUMBER NUMBER RETURN BIN_OP VAR VAR RETURN VAR IF FUNC_CALL VAR ASSIGN VAR VAR VAR VAR IF VAR NUMBER IF FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR VAR NUMBER VAR WHILE BIN_OP VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR VAR FUNC_CALL VAR VAR VAR VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR FUNC_CALL VAR VAR VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING IF VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR IF FUNC_CALL VAR VAR VAR VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP NUMBER BIN_OP VAR VAR IF FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR IF FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys. The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank. There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline. What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0. -----Input----- The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys. -----Output----- Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1. -----Examples----- Input 6 9 2 4 Output 4 Input 6 10 2 4 Output 2 Input 6 5 4 3 Output -1 -----Note----- In the first example the bus needs to refuel during each journey. In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty. In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.	a, b, f, k = map(int, input().split()) now = b da = [0, f, a] an = 0 for r in range(k): if r == k - 1: if r % 2 == 0: if now < f: print(-1) exit(0) now -= f if now < a - f: now = b an += 1 if now < a - f: print(-1) exit(0) now -= a - f else: if now < a - f: print(-1) exit(0) now -= a - f if now < f: now = b an += 1 if now < f: print(-1) exit(0) now -= f elif r % 2 == 0: if now < f: print(-1) exit(0) now -= f if now < 2 * (a - f): now = b an += 1 if now < a - f: print(-1) exit(0) now -= a - f else: if now < a - f: print(-1) exit(0) now -= a - f if now < 2 * f: now = b an += 1 if now < f: print(-1) exit(0) now -= f print(an)	ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR LIST NUMBER VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR BIN_OP VAR NUMBER IF BIN_OP VAR NUMBER NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER VAR VAR IF VAR BIN_OP VAR VAR ASSIGN VAR VAR VAR NUMBER IF VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER VAR BIN_OP VAR VAR IF VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER VAR BIN_OP VAR VAR IF VAR VAR ASSIGN VAR VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER VAR VAR IF BIN_OP VAR NUMBER NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER VAR VAR IF VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR VAR VAR NUMBER IF VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER VAR BIN_OP VAR VAR IF VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER VAR BIN_OP VAR VAR IF VAR BIN_OP NUMBER VAR ASSIGN VAR VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys. The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank. There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline. What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0. -----Input----- The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys. -----Output----- Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1. -----Examples----- Input 6 9 2 4 Output 4 Input 6 10 2 4 Output 2 Input 6 5 4 3 Output -1 -----Note----- In the first example the bus needs to refuel during each journey. In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty. In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.	a, b, f, k = list(map(int, input().split())) t = b ans = 0 def go(dist): global ans, t if dist > b: print(-1) exit() if t < dist: t = b ans += 1 t -= dist go(f) fw = True for _ in range(k - 1): go(2 * (a - f if fw else f)) fw = not fw go(a - f if fw else f) print(ans)	ASSIGN VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR NUMBER FUNC_DEF IF VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR IF VAR VAR ASSIGN VAR VAR VAR NUMBER VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP NUMBER VAR BIN_OP VAR VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR VAR
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys. The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank. There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline. What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0. -----Input----- The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys. -----Output----- Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1. -----Examples----- Input 6 9 2 4 Output 4 Input 6 10 2 4 Output 2 Input 6 5 4 3 Output -1 -----Note----- In the first example the bus needs to refuel during each journey. In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty. In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.	def list_input(): return list(map(int, input().split())) def map_input(): return map(int, input().split()) def map_string(): return input().split() a, b, f, k = map_input() tot = a * k s = 2 * a - f cur = 0 cnt = b go = 0 ans = 0 while cur < tot: go = 1 - go if go == 1: if cnt < s and cnt < tot - cur: if cnt < f: print(-1) break cnt = b ans += 1 cnt -= a - f else: cnt -= a elif cnt < a + f and cnt < tot - cur: if cnt < a - f: print(-1) break cnt = b ans += 1 cnt -= f else: cnt -= a cur += a if cnt < 0: print(-1) break else: print(ans)	FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR VAR FUNC_CALL VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP BIN_OP NUMBER VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR ASSIGN VAR BIN_OP NUMBER VAR IF VAR NUMBER IF VAR VAR VAR BIN_OP VAR VAR IF VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR VAR VAR NUMBER VAR BIN_OP VAR VAR VAR VAR IF VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR IF VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER ASSIGN VAR VAR VAR NUMBER VAR VAR VAR VAR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys. The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank. There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline. What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0. -----Input----- The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys. -----Output----- Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1. -----Examples----- Input 6 9 2 4 Output 4 Input 6 10 2 4 Output 2 Input 6 5 4 3 Output -1 -----Note----- In the first example the bus needs to refuel during each journey. In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty. In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.	def get_raw_input(): raw = input() raw = raw.split(" ") return raw def is_tour_possible(total, capo, pos, jrns): x = total - pos if (2 * x > capo or 2 * pos > capo) and jrns > 2: return False if jrns == 1 and (x > capo or pos > capo): return False if jrns == 2 and (2 * x > capo or pos > capo): return False return True def need_refuel(curr, next_stop_dst): if next_stop_dst > curr: return True return False def get_next_distance(iter, dst1, dst2, *special): if special.get("last") is None: if iter % 2 == 1: return dst2 return dst1 if special.get("last"): if iter % 2 == 1: return dst2 / 2 return dst1 / 2 def drive(iter): if iter % 2 == 1: return BtoC return AtoB inp = get_raw_input() total_dst = int(inp[0]) fuel_capo = int(inp[1]) fuel_pos = int(inp[2]) jrn_amount = int(inp[3]) AtoB = 2 fuel_pos BtoC = 2 * (total_dst - fuel_pos) refuels = 0 if is_tour_possible(total_dst, fuel_capo, fuel_pos, jrn_amount): jrn = 1 curr_fuel = fuel_capo - fuel_pos while jrn <= jrn_amount - 1: if need_refuel(curr_fuel, get_next_distance(jrn, AtoB, BtoC)): refuels += 1 curr_fuel = fuel_capo curr_fuel -= drive(jrn) jrn += 1 if need_refuel(curr_fuel, get_next_distance(jrn, AtoB, BtoC, last=True)): refuels += 1 print(refuels) else: print(-1)	FUNC_DEF ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR STRING RETURN VAR FUNC_DEF ASSIGN VAR BIN_OP VAR VAR IF BIN_OP NUMBER VAR VAR BIN_OP NUMBER VAR VAR VAR NUMBER RETURN NUMBER IF VAR NUMBER VAR VAR VAR VAR RETURN NUMBER IF VAR NUMBER BIN_OP NUMBER VAR VAR VAR VAR RETURN NUMBER RETURN NUMBER FUNC_DEF IF VAR VAR RETURN NUMBER RETURN NUMBER FUNC_DEF IF FUNC_CALL VAR STRING NONE IF BIN_OP VAR NUMBER NUMBER RETURN VAR RETURN VAR IF FUNC_CALL VAR STRING IF BIN_OP VAR NUMBER NUMBER RETURN BIN_OP VAR NUMBER RETURN BIN_OP VAR NUMBER FUNC_DEF IF BIN_OP VAR NUMBER NUMBER RETURN VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP NUMBER VAR ASSIGN VAR BIN_OP NUMBER BIN_OP VAR VAR ASSIGN VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR WHILE VAR BIN_OP VAR NUMBER IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR VAR NUMBER ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR NUMBER IF FUNC_CALL VAR VAR FUNC_CALL VAR VAR VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys. The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank. There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline. What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0. -----Input----- The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys. -----Output----- Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1. -----Examples----- Input 6 9 2 4 Output 4 Input 6 10 2 4 Output 2 Input 6 5 4 3 Output -1 -----Note----- In the first example the bus needs to refuel during each journey. In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty. In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.	value = input().split() finish, tank, azs, count = int(value[0]), int(value[1]), int(value[2]), int(value[3]) before_azs = (finish - azs) * 2 if count > 1 else finish - azs, finish * 2 - ( finish - azs ) * 2 tank_now = tank - azs if tank_now < 0: print(-1) exit() lent = finish * count - azs values, binar = 0, 0 while lent > 0: if tank_now - before_azs[binar] >= 0 or tank_now - lent >= 0: tank_now -= before_azs[binar] lent -= before_azs[binar] if lent <= 0: break if binar == 1: binar = 0 else: binar = 1 if tank_now - before_azs[binar] < 0 and not tank_now - lent >= 0: values += 1 tank_now = tank if before_azs[binar] > tank_now and not tank_now - lent >= 0: print(-1) exit() print(values)	ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER ASSIGN VAR VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER BIN_OP VAR VAR BIN_OP BIN_OP VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR ASSIGN VAR VAR NUMBER NUMBER WHILE VAR NUMBER IF BIN_OP VAR VAR VAR NUMBER BIN_OP VAR VAR NUMBER VAR VAR VAR VAR VAR VAR IF VAR NUMBER IF VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP VAR VAR VAR NUMBER BIN_OP VAR VAR NUMBER VAR NUMBER ASSIGN VAR VAR IF VAR VAR VAR BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys. The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank. There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline. What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0. -----Input----- The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys. -----Output----- Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1. -----Examples----- Input 6 9 2 4 Output 4 Input 6 10 2 4 Output 2 Input 6 5 4 3 Output -1 -----Note----- In the first example the bus needs to refuel during each journey. In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty. In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.	a, b, f, k = map(int, input().split()) tmp = b flag = 0 c = 0 for i in range(k): if i % 2 == 0: if f > b: flag = 1 break else: b = b - f if i != k - 1: if 2 * (a - f) > b: c += 1 b = tmp b = b - (a - f) elif a - f > b: b = tmp if a - f > b: flag = 1 else: c += 1 elif b < a - f: flag = 1 break else: b = b - (a - f) if i != k - 1: if 2 * f > b: c += 1 b = tmp b = b - f elif f > b: b = tmp if f > b: flag = 1 else: c += 1 if flag == 0: print(c) else: print(-1)	ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF BIN_OP VAR NUMBER NUMBER IF VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF VAR BIN_OP VAR NUMBER IF BIN_OP NUMBER BIN_OP VAR VAR VAR VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR IF BIN_OP VAR VAR VAR ASSIGN VAR VAR IF BIN_OP VAR VAR VAR ASSIGN VAR NUMBER VAR NUMBER IF VAR BIN_OP VAR VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR IF VAR BIN_OP VAR NUMBER IF BIN_OP NUMBER VAR VAR VAR NUMBER ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR VAR ASSIGN VAR VAR IF VAR VAR ASSIGN VAR NUMBER VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys. The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank. There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline. What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0. -----Input----- The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys. -----Output----- Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1. -----Examples----- Input 6 9 2 4 Output 4 Input 6 10 2 4 Output 2 Input 6 5 4 3 Output -1 -----Note----- In the first example the bus needs to refuel during each journey. In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty. In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.	a, b, f, k = [int(i) for i in input().split()] if b < f: print(-1) return journeys = 0 previous = 0 refuels1 = 0 tank = b - f while journeys != k: if previous == 0: if tank >= a - f + a * (k - journeys - 1): print(refuels1) return if b >= a - f + a * (k - journeys - 1): print(refuels1 + 1) return if tank >= 2 * (a - f): tank -= 2 * (a - f) elif b >= 2 * (a - f): refuels1 += 1 tank = b - 2 * (a - f) else: print(-1) return journeys += 1 previous = a if previous == a: if tank >= f + a * (k - journeys - 1): print(refuels1) return if b >= f + a * (k - journeys - 1): print(refuels1 + 1) return if tank >= 2 * f: tank -= 2 * f elif b >= 2 * f: refuels1 += 1 tank = b - 2 * f else: print(-1) return journeys += 1 previous = 0 if journeys == k: print(refuels1) else: print(-1)	ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR NUMBER RETURN ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR WHILE VAR VAR IF VAR NUMBER IF VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR VAR RETURN IF VAR BIN_OP BIN_OP VAR VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER RETURN IF VAR BIN_OP NUMBER BIN_OP VAR VAR VAR BIN_OP NUMBER BIN_OP VAR VAR IF VAR BIN_OP NUMBER BIN_OP VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP NUMBER BIN_OP VAR VAR EXPR FUNC_CALL VAR NUMBER RETURN VAR NUMBER ASSIGN VAR VAR IF VAR VAR IF VAR BIN_OP VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR VAR RETURN IF VAR BIN_OP VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER RETURN IF VAR BIN_OP NUMBER VAR VAR BIN_OP NUMBER VAR IF VAR BIN_OP NUMBER VAR VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR NUMBER RETURN VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR NUMBER
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys. The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank. There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline. What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0. -----Input----- The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys. -----Output----- Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1. -----Examples----- Input 6 9 2 4 Output 4 Input 6 10 2 4 Output 2 Input 6 5 4 3 Output -1 -----Note----- In the first example the bus needs to refuel during each journey. In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty. In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.	a, b, f, k = list(map(int, input().split())) now = b path = 0 ans = 0 for i in range(k): if path == 0: if i == k - 1 and now >= a: break now -= f if now < 0: print(-1) return if 2 * (a - f) <= now: now -= a - f else: ans += 1 now = b - (a - f) path = 1 elif path == 1: if i == k - 1 and now >= a: break now -= a - f if now < 0: print(-1) return if 2 * f <= now: now -= f else: ans += 1 now = b - f path = 0 if now < 0: print(-1) return print(ans)	ASSIGN VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR IF VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER RETURN IF BIN_OP NUMBER BIN_OP VAR VAR VAR VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR VAR VAR BIN_OP VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER RETURN IF BIN_OP NUMBER VAR VAR VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER RETURN EXPR FUNC_CALL VAR VAR
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys. The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank. There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline. What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0. -----Input----- The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys. -----Output----- Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1. -----Examples----- Input 6 9 2 4 Output 4 Input 6 10 2 4 Output 2 Input 6 5 4 3 Output -1 -----Note----- In the first example the bus needs to refuel during each journey. In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty. In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.	a, b, f, k = map(lambda x: int(x), input().split()) if b < f: print(-1) exit() cnt = 0 t = b - f g = f = a - f for i in range(k): if t < 0: break if k - i - 1: g += f if t < g: cnt += 1 t = b t -= g g = f = a - f print(-1 if t < 0 else cnt)	ASSIGN VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR VAR IF VAR NUMBER IF BIN_OP BIN_OP VAR VAR NUMBER VAR VAR IF VAR VAR VAR NUMBER ASSIGN VAR VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR VAR NUMBER NUMBER VAR
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys. The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank. There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline. What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0. -----Input----- The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys. -----Output----- Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1. -----Examples----- Input 6 9 2 4 Output 4 Input 6 10 2 4 Output 2 Input 6 5 4 3 Output -1 -----Note----- In the first example the bus needs to refuel during each journey. In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty. In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.	a, b, f, k = map(int, input().split()) start = True tank = b refuels = 0 possible = True while k > 0 and possible: if k == 1: if tank >= a: k -= 1 continue if start: if tank >= f and b >= a - f: refuels += 1 else: possible = False elif tank >= a - f and b >= f: refuels += 1 else: possible = False else: if start: if tank < f: possible = False elif tank >= a + (a - f): tank -= a else: tank = b - (a - f) refuels += 1 elif tank < a - f: possible = False elif tank >= a + f: tank -= a else: tank = b - f refuels += 1 start = not start k -= 1 if not possible: print(-1) else: print(refuels)	ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR NUMBER VAR IF VAR NUMBER IF VAR VAR VAR NUMBER IF VAR IF VAR VAR VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR BIN_OP VAR VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER IF VAR IF VAR VAR ASSIGN VAR NUMBER IF VAR BIN_OP VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR VAR NUMBER IF VAR BIN_OP VAR VAR ASSIGN VAR NUMBER IF VAR BIN_OP VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR VAR VAR NUMBER IF VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys. The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank. There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline. What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0. -----Input----- The first line contains four integers a, b, f, k (0 < f < a ≤ 10^6, 1 ≤ b ≤ 10^9, 1 ≤ k ≤ 10^4) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys. -----Output----- Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1. -----Examples----- Input 6 9 2 4 Output 4 Input 6 10 2 4 Output 2 Input 6 5 4 3 Output -1 -----Note----- In the first example the bus needs to refuel during each journey. In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty. In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.	from sys import exit a, b, f, k = map(int, input().split()) item, res = b, 0 if item < f: print(-1) exit() item -= f x, y = f * 2, (a - f) * 2 for i in range(1, k): if i % 2 == 1: if item < y: res += 1 item = b - y else: item -= y if item < 0: print(-1) exit() else: if item < x: res += 1 item = b - x else: item -= x if item < 0: print(-1) exit() if k % 2 == 1: if item < a - f: res += 1 item = b - (a - f) else: item -= a - f if item < 0: print(-1) exit() else: if item < f: res += 1 item = b - f else: item -= f if item < 0: print(-1) exit() print(res)	ASSIGN VAR VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER VAR IF BIN_OP VAR NUMBER NUMBER IF VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR IF VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR IF BIN_OP VAR NUMBER NUMBER IF VAR BIN_OP VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP VAR VAR VAR BIN_OP VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR IF VAR VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR VAR VAR IF VAR NUMBER EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR VAR
You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor. You must paint a fence which consists of $10^{100}$ planks in two colors in the following way (suppose planks are numbered from left to right from $0$): if the index of the plank is divisible by $r$ (such planks have indices $0$, $r$, $2r$ and so on) then you must paint it red; if the index of the plank is divisible by $b$ (such planks have indices $0$, $b$, $2b$ and so on) then you must paint it blue; if the index is divisible both by $r$ and $b$ you can choose the color to paint the plank; otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it). Furthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are $k$ consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed. The question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs. -----Input----- The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. The next $T$ lines contain descriptions of test cases — one per line. Each test case contains three integers $r$, $b$, $k$ ($1 \le r, b \le 10^9$, $2 \le k \le 10^9$) — the corresponding coefficients. -----Output----- Print $T$ words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise. -----Example----- Input 4 1 1 2 2 10 4 5 2 3 3 2 2 Output OBEY REBEL OBEY OBEY	def gcd(a, c): while a and c: if a > c: a, c = c, a c %= a return a + c for i in range(int(input())): r, b, k = map(int, input().split()) if r > b: r, b = b, r g = gcd(r, b) r, b = r // g, b // g print("REBEL" if (k - 1) * r + 1 < b else "OBEY")	FUNC_DEF WHILE VAR VAR IF VAR VAR ASSIGN VAR VAR VAR VAR VAR VAR RETURN BIN_OP VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR NUMBER VAR STRING STRING
You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor. You must paint a fence which consists of $10^{100}$ planks in two colors in the following way (suppose planks are numbered from left to right from $0$): if the index of the plank is divisible by $r$ (such planks have indices $0$, $r$, $2r$ and so on) then you must paint it red; if the index of the plank is divisible by $b$ (such planks have indices $0$, $b$, $2b$ and so on) then you must paint it blue; if the index is divisible both by $r$ and $b$ you can choose the color to paint the plank; otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it). Furthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are $k$ consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed. The question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs. -----Input----- The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. The next $T$ lines contain descriptions of test cases — one per line. Each test case contains three integers $r$, $b$, $k$ ($1 \le r, b \le 10^9$, $2 \le k \le 10^9$) — the corresponding coefficients. -----Output----- Print $T$ words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise. -----Example----- Input 4 1 1 2 2 10 4 5 2 3 3 2 2 Output OBEY REBEL OBEY OBEY	def gcd(a, b): if b == 0: return a else: return gcd(b, a % b) n = int(input()) for _ in range(n): a, b, k = map(int, input().split()) d = gcd(a, b) a = a // d b = b // d mx = max(a, b) mn = min(a, b) if mn * (k - 1) >= mx - 1: print("OBEY") else: print("REBEL")	FUNC_DEF IF VAR NUMBER RETURN VAR RETURN FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF BIN_OP VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor. You must paint a fence which consists of $10^{100}$ planks in two colors in the following way (suppose planks are numbered from left to right from $0$): if the index of the plank is divisible by $r$ (such planks have indices $0$, $r$, $2r$ and so on) then you must paint it red; if the index of the plank is divisible by $b$ (such planks have indices $0$, $b$, $2b$ and so on) then you must paint it blue; if the index is divisible both by $r$ and $b$ you can choose the color to paint the plank; otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it). Furthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are $k$ consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed. The question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs. -----Input----- The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. The next $T$ lines contain descriptions of test cases — one per line. Each test case contains three integers $r$, $b$, $k$ ($1 \le r, b \le 10^9$, $2 \le k \le 10^9$) — the corresponding coefficients. -----Output----- Print $T$ words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise. -----Example----- Input 4 1 1 2 2 10 4 5 2 3 3 2 2 Output OBEY REBEL OBEY OBEY	def gcd(a, b): return a if b == 0 else gcd(b, a % b) t = int(input()) for case_num in range(t): r, b, k = map(int, input().split(" ")) if r > b: tmp = r r = b b = tmp g = gcd(r, b) b = b // g r = r // g n = (b - 2) // r + 1 if b % r != 0 else b // r - 1 print("REBEL" if n >= k else "OBEY")	FUNC_DEF RETURN VAR NUMBER VAR FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING IF VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR NUMBER BIN_OP BIN_OP BIN_OP VAR NUMBER VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR VAR VAR STRING STRING
You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor. You must paint a fence which consists of $10^{100}$ planks in two colors in the following way (suppose planks are numbered from left to right from $0$): if the index of the plank is divisible by $r$ (such planks have indices $0$, $r$, $2r$ and so on) then you must paint it red; if the index of the plank is divisible by $b$ (such planks have indices $0$, $b$, $2b$ and so on) then you must paint it blue; if the index is divisible both by $r$ and $b$ you can choose the color to paint the plank; otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it). Furthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are $k$ consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed. The question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs. -----Input----- The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. The next $T$ lines contain descriptions of test cases — one per line. Each test case contains three integers $r$, $b$, $k$ ($1 \le r, b \le 10^9$, $2 \le k \le 10^9$) — the corresponding coefficients. -----Output----- Print $T$ words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise. -----Example----- Input 4 1 1 2 2 10 4 5 2 3 3 2 2 Output OBEY REBEL OBEY OBEY	n = int(input()) def gcd(x, y): while x != 0 and y != 0: x, y = max(x, y), min(x, y) x %= y return max(x, y) for i in range(n): r, b, k = list(map(int, input().split())) g = gcd(r, b) r, b = max(r, b) // g, min(r, b) // g if ( r % b == 0 and r // b - 1 < k or r % b == 1 and r // b < k or r % b > 1 and r // b + 1 < k ): print("OBEY") else: print("REBEL")	ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF WHILE VAR NUMBER VAR NUMBER ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR VAR VAR RETURN FUNC_CALL VAR VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR BIN_OP FUNC_CALL VAR VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR VAR IF BIN_OP VAR VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER VAR BIN_OP VAR VAR NUMBER BIN_OP VAR VAR VAR BIN_OP VAR VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor. You must paint a fence which consists of $10^{100}$ planks in two colors in the following way (suppose planks are numbered from left to right from $0$): if the index of the plank is divisible by $r$ (such planks have indices $0$, $r$, $2r$ and so on) then you must paint it red; if the index of the plank is divisible by $b$ (such planks have indices $0$, $b$, $2b$ and so on) then you must paint it blue; if the index is divisible both by $r$ and $b$ you can choose the color to paint the plank; otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it). Furthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are $k$ consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed. The question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs. -----Input----- The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. The next $T$ lines contain descriptions of test cases — one per line. Each test case contains three integers $r$, $b$, $k$ ($1 \le r, b \le 10^9$, $2 \le k \le 10^9$) — the corresponding coefficients. -----Output----- Print $T$ words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise. -----Example----- Input 4 1 1 2 2 10 4 5 2 3 3 2 2 Output OBEY REBEL OBEY OBEY	t = int(input()) for i in range(t): r, b, k = list(map(int, input().split())) if r > b: temp = r r = b b = temp a = b // r c = b % r if c == 0: a -= 1 if c > 0 and r % c > 0: a += 1 if a >= k: print("REBEL") else: print("OBEY")	ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR IF VAR NUMBER VAR NUMBER IF VAR NUMBER BIN_OP VAR VAR NUMBER VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor. You must paint a fence which consists of $10^{100}$ planks in two colors in the following way (suppose planks are numbered from left to right from $0$): if the index of the plank is divisible by $r$ (such planks have indices $0$, $r$, $2r$ and so on) then you must paint it red; if the index of the plank is divisible by $b$ (such planks have indices $0$, $b$, $2b$ and so on) then you must paint it blue; if the index is divisible both by $r$ and $b$ you can choose the color to paint the plank; otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it). Furthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are $k$ consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed. The question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs. -----Input----- The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. The next $T$ lines contain descriptions of test cases — one per line. Each test case contains three integers $r$, $b$, $k$ ($1 \le r, b \le 10^9$, $2 \le k \le 10^9$) — the corresponding coefficients. -----Output----- Print $T$ words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise. -----Example----- Input 4 1 1 2 2 10 4 5 2 3 3 2 2 Output OBEY REBEL OBEY OBEY	t = int(input()) def gcd(x, y): while y % x != 0: r = y % x x, y = r, x return x for i in range(t): r, b, k = list(map(int, input().split())) l = max(r, b) s = min(r, b) d = gcd(s, l) x = l // d - 1 if (x + s // d - 1) // (s // d) >= k: print("REBEL") else: print("OBEY")	ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF WHILE BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR VAR VAR RETURN VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER IF BIN_OP BIN_OP BIN_OP VAR BIN_OP VAR VAR NUMBER BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor. You must paint a fence which consists of $10^{100}$ planks in two colors in the following way (suppose planks are numbered from left to right from $0$): if the index of the plank is divisible by $r$ (such planks have indices $0$, $r$, $2r$ and so on) then you must paint it red; if the index of the plank is divisible by $b$ (such planks have indices $0$, $b$, $2b$ and so on) then you must paint it blue; if the index is divisible both by $r$ and $b$ you can choose the color to paint the plank; otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it). Furthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are $k$ consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed. The question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs. -----Input----- The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. The next $T$ lines contain descriptions of test cases — one per line. Each test case contains three integers $r$, $b$, $k$ ($1 \le r, b \le 10^9$, $2 \le k \le 10^9$) — the corresponding coefficients. -----Output----- Print $T$ words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise. -----Example----- Input 4 1 1 2 2 10 4 5 2 3 3 2 2 Output OBEY REBEL OBEY OBEY	t = int(input()) def gwc(r, b): while r % b: r, b = b, r % b return b for j in range(t): r, b, k = [int(i) for i in input().split()] if r < b: r, b = b, r c = gwc(r, b) r, b = r // c, b // c if r % b == 0 and r // b <= k: print("OBEY") elif r > b * (k - 1) + 1: print("REBEL") else: print("OBEY")	ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF WHILE BIN_OP VAR VAR ASSIGN VAR VAR VAR BIN_OP VAR VAR RETURN VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR BIN_OP VAR VAR BIN_OP VAR VAR IF BIN_OP VAR VAR NUMBER BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR STRING IF VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor. You must paint a fence which consists of $10^{100}$ planks in two colors in the following way (suppose planks are numbered from left to right from $0$): if the index of the plank is divisible by $r$ (such planks have indices $0$, $r$, $2r$ and so on) then you must paint it red; if the index of the plank is divisible by $b$ (such planks have indices $0$, $b$, $2b$ and so on) then you must paint it blue; if the index is divisible both by $r$ and $b$ you can choose the color to paint the plank; otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it). Furthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are $k$ consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed. The question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs. -----Input----- The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. The next $T$ lines contain descriptions of test cases — one per line. Each test case contains three integers $r$, $b$, $k$ ($1 \le r, b \le 10^9$, $2 \le k \le 10^9$) — the corresponding coefficients. -----Output----- Print $T$ words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise. -----Example----- Input 4 1 1 2 2 10 4 5 2 3 3 2 2 Output OBEY REBEL OBEY OBEY	def gcd(a, b): if b == 0: return a else: return gcd(b, a % b) t = int(input()) for i in range(t): b, r, k = map(int, input().split()) if (max(b, r) - gcd(b, r) + min(b, r) - 1) // min(b, r) < k: print("OBEY") else: print("REBEL")	FUNC_DEF IF VAR NUMBER RETURN VAR RETURN FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF BIN_OP BIN_OP BIN_OP BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR NUMBER FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor. You must paint a fence which consists of $10^{100}$ planks in two colors in the following way (suppose planks are numbered from left to right from $0$): if the index of the plank is divisible by $r$ (such planks have indices $0$, $r$, $2r$ and so on) then you must paint it red; if the index of the plank is divisible by $b$ (such planks have indices $0$, $b$, $2b$ and so on) then you must paint it blue; if the index is divisible both by $r$ and $b$ you can choose the color to paint the plank; otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it). Furthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are $k$ consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed. The question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs. -----Input----- The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. The next $T$ lines contain descriptions of test cases — one per line. Each test case contains three integers $r$, $b$, $k$ ($1 \le r, b \le 10^9$, $2 \le k \le 10^9$) — the corresponding coefficients. -----Output----- Print $T$ words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise. -----Example----- Input 4 1 1 2 2 10 4 5 2 3 3 2 2 Output OBEY REBEL OBEY OBEY	def gcd(a, b): if a < b: a, b = b, a while a % b != 0: a, b = b, a % b return b w = int(input()) for i in range(0, w): r, b, k = map(int, input().split()) s = gcd(r, b) if s == b: if r / b <= k: print("OBEY") else: print("REBEL") elif s == r: if b / r <= k: print("OBEY") else: print("REBEL") else: y = int((max(r, b) - 1) / min(r, b)) if y + 1 <= k and y == (max(r, b) - gcd(r, b)) / min(r, b): print("OBEY") elif y + 1 < k and y != (max(r, b) - gcd(r, b)) / min(r, b): print("OBEY") else: print("REBEL")	FUNC_DEF IF VAR VAR ASSIGN VAR VAR VAR VAR WHILE BIN_OP VAR VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR VAR IF BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING IF VAR VAR IF BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR NUMBER FUNC_CALL VAR VAR VAR IF BIN_OP VAR NUMBER VAR VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR STRING IF BIN_OP VAR NUMBER VAR VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor. You must paint a fence which consists of $10^{100}$ planks in two colors in the following way (suppose planks are numbered from left to right from $0$): if the index of the plank is divisible by $r$ (such planks have indices $0$, $r$, $2r$ and so on) then you must paint it red; if the index of the plank is divisible by $b$ (such planks have indices $0$, $b$, $2b$ and so on) then you must paint it blue; if the index is divisible both by $r$ and $b$ you can choose the color to paint the plank; otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it). Furthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are $k$ consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed. The question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs. -----Input----- The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. The next $T$ lines contain descriptions of test cases — one per line. Each test case contains three integers $r$, $b$, $k$ ($1 \le r, b \le 10^9$, $2 \le k \le 10^9$) — the corresponding coefficients. -----Output----- Print $T$ words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise. -----Example----- Input 4 1 1 2 2 10 4 5 2 3 3 2 2 Output OBEY REBEL OBEY OBEY	def gcd(a: int, b: int) -> int: if b == 0: return a return gcd(b, a % b) t = int(input()) for _ in range(t): a, b, k = map(int, input().split()) if a > b: a, b = b, a gcd_ab = gcd(a, b) tmp = a % gcd_ab + gcd_ab if -(-(b - tmp) // a) < k: print("OBEY") else: print("REBEL")	FUNC_DEF VAR VAR IF VAR NUMBER RETURN VAR RETURN FUNC_CALL VAR VAR BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR IF BIN_OP BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor. You must paint a fence which consists of $10^{100}$ planks in two colors in the following way (suppose planks are numbered from left to right from $0$): if the index of the plank is divisible by $r$ (such planks have indices $0$, $r$, $2r$ and so on) then you must paint it red; if the index of the plank is divisible by $b$ (such planks have indices $0$, $b$, $2b$ and so on) then you must paint it blue; if the index is divisible both by $r$ and $b$ you can choose the color to paint the plank; otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it). Furthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are $k$ consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed. The question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs. -----Input----- The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. The next $T$ lines contain descriptions of test cases — one per line. Each test case contains three integers $r$, $b$, $k$ ($1 \le r, b \le 10^9$, $2 \le k \le 10^9$) — the corresponding coefficients. -----Output----- Print $T$ words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise. -----Example----- Input 4 1 1 2 2 10 4 5 2 3 3 2 2 Output OBEY REBEL OBEY OBEY	def getGCD(a, b): if a == 0 or b == 0: return max(a, b) if a > b: return getGCD(a % b, b) return getGCD(b % a, a) n = int(input()) for i in range(n): r, b, k = map(int, input().split()) bCopy = b b = max(b, r) r = min(r, bCopy) if (b - getGCD(r, b) - 1) // r + 1 >= k: print("REBEL") else: print("OBEY")	FUNC_DEF IF VAR NUMBER VAR NUMBER RETURN FUNC_CALL VAR VAR VAR IF VAR VAR RETURN FUNC_CALL VAR BIN_OP VAR VAR VAR RETURN FUNC_CALL VAR BIN_OP VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF BIN_OP BIN_OP BIN_OP BIN_OP VAR FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor. You must paint a fence which consists of $10^{100}$ planks in two colors in the following way (suppose planks are numbered from left to right from $0$): if the index of the plank is divisible by $r$ (such planks have indices $0$, $r$, $2r$ and so on) then you must paint it red; if the index of the plank is divisible by $b$ (such planks have indices $0$, $b$, $2b$ and so on) then you must paint it blue; if the index is divisible both by $r$ and $b$ you can choose the color to paint the plank; otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it). Furthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are $k$ consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed. The question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs. -----Input----- The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. The next $T$ lines contain descriptions of test cases — one per line. Each test case contains three integers $r$, $b$, $k$ ($1 \le r, b \le 10^9$, $2 \le k \le 10^9$) — the corresponding coefficients. -----Output----- Print $T$ words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise. -----Example----- Input 4 1 1 2 2 10 4 5 2 3 3 2 2 Output OBEY REBEL OBEY OBEY	def hcf(a, b): if a % b == 0: return b return hcf(b, a % b) def lcm(a, b): return a * b // hcf(a, b) t = int(input()) for i in range(t): a, b, k = map(int, input().split()) p = lcm(a, b) q = hcf(a, b) if k >= p: print("OBEY") elif a == b: print("OBEY") else: g = a l = b if b > a: g = b l = a if (g - q) % l == 0 and (g - q) // l < k: print("OBEY") elif (g - q) % l != 0 and (g - q) // l + 1 < k: print("OBEY") else: print("REBEL")	FUNC_DEF IF BIN_OP VAR VAR NUMBER RETURN VAR RETURN FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_DEF RETURN BIN_OP BIN_OP VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR STRING IF VAR VAR EXPR FUNC_CALL VAR STRING ASSIGN VAR VAR ASSIGN VAR VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR IF BIN_OP BIN_OP VAR VAR VAR NUMBER BIN_OP BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR STRING IF BIN_OP BIN_OP VAR VAR VAR NUMBER BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor. You must paint a fence which consists of $10^{100}$ planks in two colors in the following way (suppose planks are numbered from left to right from $0$): if the index of the plank is divisible by $r$ (such planks have indices $0$, $r$, $2r$ and so on) then you must paint it red; if the index of the plank is divisible by $b$ (such planks have indices $0$, $b$, $2b$ and so on) then you must paint it blue; if the index is divisible both by $r$ and $b$ you can choose the color to paint the plank; otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it). Furthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are $k$ consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed. The question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs. -----Input----- The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. The next $T$ lines contain descriptions of test cases — one per line. Each test case contains three integers $r$, $b$, $k$ ($1 \le r, b \le 10^9$, $2 \le k \le 10^9$) — the corresponding coefficients. -----Output----- Print $T$ words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise. -----Example----- Input 4 1 1 2 2 10 4 5 2 3 3 2 2 Output OBEY REBEL OBEY OBEY	def gcd(x, y): while y: x, y = y, x % y return x x = eval(input()) for i in range(x): y = input().split(" ") g = gcd(int(y[0]), int(y[1])) a = min(int(y[0]), int(y[1])) / g b = max(int(y[0]), int(y[1])) / g c = int(y[2]) if (b - 1) // a + 1 > c: print("REBEL") elif (b - 1) // a + 1 == c and (b - 1) % a != 0: print("REBEL") else: print("OBEY")	FUNC_DEF WHILE VAR ASSIGN VAR VAR VAR BIN_OP VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR BIN_OP FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER FUNC_CALL VAR VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER IF BIN_OP BIN_OP BIN_OP VAR NUMBER VAR NUMBER VAR EXPR FUNC_CALL VAR STRING IF BIN_OP BIN_OP BIN_OP VAR NUMBER VAR NUMBER VAR BIN_OP BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor. You must paint a fence which consists of $10^{100}$ planks in two colors in the following way (suppose planks are numbered from left to right from $0$): if the index of the plank is divisible by $r$ (such planks have indices $0$, $r$, $2r$ and so on) then you must paint it red; if the index of the plank is divisible by $b$ (such planks have indices $0$, $b$, $2b$ and so on) then you must paint it blue; if the index is divisible both by $r$ and $b$ you can choose the color to paint the plank; otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it). Furthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are $k$ consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed. The question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs. -----Input----- The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. The next $T$ lines contain descriptions of test cases — one per line. Each test case contains three integers $r$, $b$, $k$ ($1 \le r, b \le 10^9$, $2 \le k \le 10^9$) — the corresponding coefficients. -----Output----- Print $T$ words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise. -----Example----- Input 4 1 1 2 2 10 4 5 2 3 3 2 2 Output OBEY REBEL OBEY OBEY	T = int(input()) def gcd(a, b): if a > b: a, b = b, a if b % a == 0: return a return gcd(a, b % a) for case in range(T): r, b, k = list(map(int, input().split())) if r > b: r, b = b, r if k * r < b: print("REBEL") elif (k - 1) * r >= b: print("OBEY") else: g = gcd(r, b) if g == r: print("OBEY") elif g + r * (k - 1) < b: print("REBEL") else: print("OBEY")	ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF IF VAR VAR ASSIGN VAR VAR VAR VAR IF BIN_OP VAR VAR NUMBER RETURN VAR RETURN FUNC_CALL VAR VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR ASSIGN VAR VAR VAR VAR IF BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR STRING IF BIN_OP BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR STRING IF BIN_OP VAR BIN_OP VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor. You must paint a fence which consists of $10^{100}$ planks in two colors in the following way (suppose planks are numbered from left to right from $0$): if the index of the plank is divisible by $r$ (such planks have indices $0$, $r$, $2r$ and so on) then you must paint it red; if the index of the plank is divisible by $b$ (such planks have indices $0$, $b$, $2b$ and so on) then you must paint it blue; if the index is divisible both by $r$ and $b$ you can choose the color to paint the plank; otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it). Furthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are $k$ consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed. The question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs. -----Input----- The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. The next $T$ lines contain descriptions of test cases — one per line. Each test case contains three integers $r$, $b$, $k$ ($1 \le r, b \le 10^9$, $2 \le k \le 10^9$) — the corresponding coefficients. -----Output----- Print $T$ words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise. -----Example----- Input 4 1 1 2 2 10 4 5 2 3 3 2 2 Output OBEY REBEL OBEY OBEY	def func_name(input1, input2): while input2: input1, input2 = input2, input1 % input2 return input1 counter = 0 T = int(input()) for i in range(T): temp = input().split() input1 = int(temp[counter]) input2 = int(temp[counter + 1]) input3 = int(temp[counter + 2]) if input1 > input2: temp_value = input1 input1 = input2 input2 = temp_value if input3 - 1 > (input2 - 2 * func_name(input1, input2)) / input1: print("OBEY") else: print("REBEL")	FUNC_DEF WHILE VAR ASSIGN VAR VAR VAR BIN_OP VAR VAR RETURN VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR IF BIN_OP VAR NUMBER BIN_OP BIN_OP VAR BIN_OP NUMBER FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor. You must paint a fence which consists of $10^{100}$ planks in two colors in the following way (suppose planks are numbered from left to right from $0$): if the index of the plank is divisible by $r$ (such planks have indices $0$, $r$, $2r$ and so on) then you must paint it red; if the index of the plank is divisible by $b$ (such planks have indices $0$, $b$, $2b$ and so on) then you must paint it blue; if the index is divisible both by $r$ and $b$ you can choose the color to paint the plank; otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it). Furthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are $k$ consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed. The question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs. -----Input----- The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. The next $T$ lines contain descriptions of test cases — one per line. Each test case contains three integers $r$, $b$, $k$ ($1 \le r, b \le 10^9$, $2 \le k \le 10^9$) — the corresponding coefficients. -----Output----- Print $T$ words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise. -----Example----- Input 4 1 1 2 2 10 4 5 2 3 3 2 2 Output OBEY REBEL OBEY OBEY	import sys class IoTool: DEBUG = 0 def _reader_dbg(): with open("./input.txt", "r") as f: lines = f.readlines() for l in lines: yield l.strip() def _reader_oj(): return iter(sys.stdin.read().split("\n")) reader = _reader_dbg() if DEBUG else _reader_oj() def read(): return next(IoTool.reader) input = IoTool.read def gcd(a, b): return a if b == 0 else gcd(b, a % b) def main(): r, b, k = map(int, input().split()) g = gcd(r, b) r, b = min(r, b), max(r, b) if g + (k - 1) * r < b: print("REBEL") else: print("OBEY") n = int(input()) for _ in range(n): main()	IMPORT CLASS_DEF ASSIGN VAR NUMBER FUNC_DEF FUNC_CALL VAR STRING STRING VAR ASSIGN VAR FUNC_CALL VAR FOR VAR VAR EXPR FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF RETURN FUNC_CALL VAR VAR ASSIGN VAR VAR FUNC_DEF RETURN VAR NUMBER VAR FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_DEF ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF BIN_OP VAR BIN_OP BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR
You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor. You must paint a fence which consists of $10^{100}$ planks in two colors in the following way (suppose planks are numbered from left to right from $0$): if the index of the plank is divisible by $r$ (such planks have indices $0$, $r$, $2r$ and so on) then you must paint it red; if the index of the plank is divisible by $b$ (such planks have indices $0$, $b$, $2b$ and so on) then you must paint it blue; if the index is divisible both by $r$ and $b$ you can choose the color to paint the plank; otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it). Furthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are $k$ consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed. The question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs. -----Input----- The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. The next $T$ lines contain descriptions of test cases — one per line. Each test case contains three integers $r$, $b$, $k$ ($1 \le r, b \le 10^9$, $2 \le k \le 10^9$) — the corresponding coefficients. -----Output----- Print $T$ words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise. -----Example----- Input 4 1 1 2 2 10 4 5 2 3 3 2 2 Output OBEY REBEL OBEY OBEY	import sys input = lambda: sys.stdin.readline().rstrip() def gcd(x, y): while y != 0: r = x % y x = y y = r return x T = int(input()) for _ in range(T): r, b, k = map(int, input().split()) if r < b: r, b = b, r g = gcd(r, b) r = r / g b = b / g if b * (k - 1) + 1 >= r: print("OBEY") else: print("REBEL")	IMPORT ASSIGN VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF WHILE VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR IF BIN_OP BIN_OP VAR BIN_OP VAR NUMBER NUMBER VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor. You must paint a fence which consists of $10^{100}$ planks in two colors in the following way (suppose planks are numbered from left to right from $0$): if the index of the plank is divisible by $r$ (such planks have indices $0$, $r$, $2r$ and so on) then you must paint it red; if the index of the plank is divisible by $b$ (such planks have indices $0$, $b$, $2b$ and so on) then you must paint it blue; if the index is divisible both by $r$ and $b$ you can choose the color to paint the plank; otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it). Furthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are $k$ consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed. The question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs. -----Input----- The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. The next $T$ lines contain descriptions of test cases — one per line. Each test case contains three integers $r$, $b$, $k$ ($1 \le r, b \le 10^9$, $2 \le k \le 10^9$) — the corresponding coefficients. -----Output----- Print $T$ words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise. -----Example----- Input 4 1 1 2 2 10 4 5 2 3 3 2 2 Output OBEY REBEL OBEY OBEY	n = int(input()) def gcd(a, b): if b > a: ob = b b = a a = ob if b == 0: return a return gcd(a % b, b) for i in range(0, n): ln = [int(j) for j in input().split(" ")] b = ln[0] r = ln[1] k = ln[2] if b > r: ob = b b = r r = ob pd = gcd(r, b) nm = (r - pd) // b if (r - pd) % b != 0: nm += 1 if nm >= k: print("REBEL") else: print("OBEY")	ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF IF VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR IF VAR NUMBER RETURN VAR RETURN FUNC_CALL VAR BIN_OP VAR VAR VAR FOR VAR FUNC_CALL VAR NUMBER VAR ASSIGN VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER ASSIGN VAR VAR NUMBER IF VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR VAR IF BIN_OP BIN_OP VAR VAR VAR NUMBER VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor. You must paint a fence which consists of $10^{100}$ planks in two colors in the following way (suppose planks are numbered from left to right from $0$): if the index of the plank is divisible by $r$ (such planks have indices $0$, $r$, $2r$ and so on) then you must paint it red; if the index of the plank is divisible by $b$ (such planks have indices $0$, $b$, $2b$ and so on) then you must paint it blue; if the index is divisible both by $r$ and $b$ you can choose the color to paint the plank; otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it). Furthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are $k$ consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed. The question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs. -----Input----- The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. The next $T$ lines contain descriptions of test cases — one per line. Each test case contains three integers $r$, $b$, $k$ ($1 \le r, b \le 10^9$, $2 \le k \le 10^9$) — the corresponding coefficients. -----Output----- Print $T$ words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise. -----Example----- Input 4 1 1 2 2 10 4 5 2 3 3 2 2 Output OBEY REBEL OBEY OBEY	def gcd(a, b): if b == 0: return a return gcd(b, a % b) for i in " " * int(input()): r, b, k = map(int, input().split()) g = gcd(r, b) r /= g b /= g if r > b: r, b = b, r if r == 1: if b > k: print("REBEL") else: print("OBEY") elif r * (k - 1) + 1 <= b - 1: print("REBEL") else: print("OBEY")	FUNC_DEF IF VAR NUMBER RETURN VAR RETURN FUNC_CALL VAR VAR BIN_OP VAR VAR FOR VAR BIN_OP STRING FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR IF VAR VAR ASSIGN VAR VAR VAR VAR IF VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING IF BIN_OP BIN_OP VAR BIN_OP VAR NUMBER NUMBER BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor. You must paint a fence which consists of $10^{100}$ planks in two colors in the following way (suppose planks are numbered from left to right from $0$): if the index of the plank is divisible by $r$ (such planks have indices $0$, $r$, $2r$ and so on) then you must paint it red; if the index of the plank is divisible by $b$ (such planks have indices $0$, $b$, $2b$ and so on) then you must paint it blue; if the index is divisible both by $r$ and $b$ you can choose the color to paint the plank; otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it). Furthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are $k$ consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed. The question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs. -----Input----- The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. The next $T$ lines contain descriptions of test cases — one per line. Each test case contains three integers $r$, $b$, $k$ ($1 \le r, b \le 10^9$, $2 \le k \le 10^9$) — the corresponding coefficients. -----Output----- Print $T$ words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise. -----Example----- Input 4 1 1 2 2 10 4 5 2 3 3 2 2 Output OBEY REBEL OBEY OBEY	def gcd(x, y): while y: x, y = y, x % y return x t = int(input()) for nt in range(t): r, b, k = map(int, input().split()) r, b = max(r, b), min(r, b) if r == b: print("OBEY") elif (r - 1) // b >= k: print("REBEL") else: hcf = gcd(r, b) if (r - 1 + (b - hcf)) // b >= k: print("REBEL") else: print("OBEY")	FUNC_DEF WHILE VAR ASSIGN VAR VAR VAR BIN_OP VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR STRING IF BIN_OP BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR VAR IF BIN_OP BIN_OP BIN_OP VAR NUMBER BIN_OP VAR VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor. You must paint a fence which consists of $10^{100}$ planks in two colors in the following way (suppose planks are numbered from left to right from $0$): if the index of the plank is divisible by $r$ (such planks have indices $0$, $r$, $2r$ and so on) then you must paint it red; if the index of the plank is divisible by $b$ (such planks have indices $0$, $b$, $2b$ and so on) then you must paint it blue; if the index is divisible both by $r$ and $b$ you can choose the color to paint the plank; otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it). Furthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are $k$ consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed. The question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs. -----Input----- The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. The next $T$ lines contain descriptions of test cases — one per line. Each test case contains three integers $r$, $b$, $k$ ($1 \le r, b \le 10^9$, $2 \le k \le 10^9$) — the corresponding coefficients. -----Output----- Print $T$ words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise. -----Example----- Input 4 1 1 2 2 10 4 5 2 3 3 2 2 Output OBEY REBEL OBEY OBEY	t = int(input()) def ext_gcd(p, q): if q == 0: return p, 1, 0 g, y, x = ext_gcd(q, p % q) y -= p // q * x return g, x, y for _ in range(t): r, b, k = map(int, input().split()) ponta = ext_gcd(r, b)[0] if ponta <= b - 1 - (k - 1) * r or ponta <= r - 1 - (k - 1) * b: print("REBEL") else: print("OBEY")	ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF IF VAR NUMBER RETURN VAR NUMBER NUMBER ASSIGN VAR VAR VAR FUNC_CALL VAR VAR BIN_OP VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR RETURN VAR VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR NUMBER IF VAR BIN_OP BIN_OP VAR NUMBER BIN_OP BIN_OP VAR NUMBER VAR VAR BIN_OP BIN_OP VAR NUMBER BIN_OP BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor. You must paint a fence which consists of $10^{100}$ planks in two colors in the following way (suppose planks are numbered from left to right from $0$): if the index of the plank is divisible by $r$ (such planks have indices $0$, $r$, $2r$ and so on) then you must paint it red; if the index of the plank is divisible by $b$ (such planks have indices $0$, $b$, $2b$ and so on) then you must paint it blue; if the index is divisible both by $r$ and $b$ you can choose the color to paint the plank; otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it). Furthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are $k$ consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed. The question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs. -----Input----- The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. The next $T$ lines contain descriptions of test cases — one per line. Each test case contains three integers $r$, $b$, $k$ ($1 \le r, b \le 10^9$, $2 \le k \le 10^9$) — the corresponding coefficients. -----Output----- Print $T$ words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise. -----Example----- Input 4 1 1 2 2 10 4 5 2 3 3 2 2 Output OBEY REBEL OBEY OBEY	import sys input = sys.stdin.readline def gcd(a, b): if b == 0: return a else: return gcd(b, a % b) t = int(input()) for testcase in range(t): r, b, k = map(int, input().split()) if r == b: print("OBEY") continue if r < b: r, b = b, r g = gcd(r, b) if r > (k - 1) * b + g: print("REBEL") else: print("OBEY")	IMPORT ASSIGN VAR VAR FUNC_DEF IF VAR NUMBER RETURN VAR RETURN FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR STRING IF VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor. You must paint a fence which consists of $10^{100}$ planks in two colors in the following way (suppose planks are numbered from left to right from $0$): if the index of the plank is divisible by $r$ (such planks have indices $0$, $r$, $2r$ and so on) then you must paint it red; if the index of the plank is divisible by $b$ (such planks have indices $0$, $b$, $2b$ and so on) then you must paint it blue; if the index is divisible both by $r$ and $b$ you can choose the color to paint the plank; otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it). Furthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are $k$ consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed. The question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs. -----Input----- The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. The next $T$ lines contain descriptions of test cases — one per line. Each test case contains three integers $r$, $b$, $k$ ($1 \le r, b \le 10^9$, $2 \le k \le 10^9$) — the corresponding coefficients. -----Output----- Print $T$ words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise. -----Example----- Input 4 1 1 2 2 10 4 5 2 3 3 2 2 Output OBEY REBEL OBEY OBEY	def gcd(a, b): while b != 0: temp = a % b a = b b = temp return abs(a) for _ in range(int(input())): r, b, k = map(int, input().split()) r, b = min(r, b), max(r, b) g = gcd(r, b) r //= g b //= g x = (b - 2) // r + 1 if x >= k: print("REBEL") else: print("OBEY")	FUNC_DEF WHILE VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR RETURN FUNC_CALL VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor. You must paint a fence which consists of $10^{100}$ planks in two colors in the following way (suppose planks are numbered from left to right from $0$): if the index of the plank is divisible by $r$ (such planks have indices $0$, $r$, $2r$ and so on) then you must paint it red; if the index of the plank is divisible by $b$ (such planks have indices $0$, $b$, $2b$ and so on) then you must paint it blue; if the index is divisible both by $r$ and $b$ you can choose the color to paint the plank; otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it). Furthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are $k$ consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed. The question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs. -----Input----- The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. The next $T$ lines contain descriptions of test cases — one per line. Each test case contains three integers $r$, $b$, $k$ ($1 \le r, b \le 10^9$, $2 \le k \le 10^9$) — the corresponding coefficients. -----Output----- Print $T$ words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise. -----Example----- Input 4 1 1 2 2 10 4 5 2 3 3 2 2 Output OBEY REBEL OBEY OBEY	def gcd(a, b): while b: a, b = b, a % b return abs(a) N = int(input()) for _ in range(N): a, b, k = list(map(int, input().split())) g = gcd(a, b) mi = min(a, b) ma = max(a, b) A = ma // g - 1 B = mi // g if B * (k - 1) < A: print("REBEL") else: print("OBEY")	FUNC_DEF WHILE VAR ASSIGN VAR VAR VAR BIN_OP VAR VAR RETURN FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR IF BIN_OP VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor. You must paint a fence which consists of $10^{100}$ planks in two colors in the following way (suppose planks are numbered from left to right from $0$): if the index of the plank is divisible by $r$ (such planks have indices $0$, $r$, $2r$ and so on) then you must paint it red; if the index of the plank is divisible by $b$ (such planks have indices $0$, $b$, $2b$ and so on) then you must paint it blue; if the index is divisible both by $r$ and $b$ you can choose the color to paint the plank; otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it). Furthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are $k$ consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed. The question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs. -----Input----- The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. The next $T$ lines contain descriptions of test cases — one per line. Each test case contains three integers $r$, $b$, $k$ ($1 \le r, b \le 10^9$, $2 \le k \le 10^9$) — the corresponding coefficients. -----Output----- Print $T$ words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise. -----Example----- Input 4 1 1 2 2 10 4 5 2 3 3 2 2 Output OBEY REBEL OBEY OBEY	def gcd(a, b): if a < b: a, b = b, a while b: a, b = b, a % b return a def f(): r, b, k = [int(s) for s in input().split()] if r > b: r, b = b, r g = gcd(r, b) r //= g b //= g rangeLen = r * (k - 1) + 1 if rangeLen <= b - 1: print("REBEL") else: print("OBEY") n = int(input()) for i in range(n): f()	FUNC_DEF IF VAR VAR ASSIGN VAR VAR VAR VAR WHILE VAR ASSIGN VAR VAR VAR BIN_OP VAR VAR RETURN VAR FUNC_DEF ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER NUMBER IF VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR
You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor. You must paint a fence which consists of $10^{100}$ planks in two colors in the following way (suppose planks are numbered from left to right from $0$): if the index of the plank is divisible by $r$ (such planks have indices $0$, $r$, $2r$ and so on) then you must paint it red; if the index of the plank is divisible by $b$ (such planks have indices $0$, $b$, $2b$ and so on) then you must paint it blue; if the index is divisible both by $r$ and $b$ you can choose the color to paint the plank; otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it). Furthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are $k$ consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed. The question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs. -----Input----- The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. The next $T$ lines contain descriptions of test cases — one per line. Each test case contains three integers $r$, $b$, $k$ ($1 \le r, b \le 10^9$, $2 \le k \le 10^9$) — the corresponding coefficients. -----Output----- Print $T$ words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise. -----Example----- Input 4 1 1 2 2 10 4 5 2 3 3 2 2 Output OBEY REBEL OBEY OBEY	t = int(input()) def gcd(a, b): while not (a == 0 or b == 0): if a < b: a, b = b, a a %= b return a + b for i in range(t): r, b, k = map(int, input().split()) g = gcd(r, b) if r > b: r, b = b, r r //= g b //= g val = (k - 1) * r + 1 if val < b: print("REBEL") else: print("OBEY")	ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF WHILE VAR NUMBER VAR NUMBER IF VAR VAR ASSIGN VAR VAR VAR VAR VAR VAR RETURN BIN_OP VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR VAR ASSIGN VAR VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor. You must paint a fence which consists of $10^{100}$ planks in two colors in the following way (suppose planks are numbered from left to right from $0$): if the index of the plank is divisible by $r$ (such planks have indices $0$, $r$, $2r$ and so on) then you must paint it red; if the index of the plank is divisible by $b$ (such planks have indices $0$, $b$, $2b$ and so on) then you must paint it blue; if the index is divisible both by $r$ and $b$ you can choose the color to paint the plank; otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it). Furthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are $k$ consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed. The question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs. -----Input----- The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. The next $T$ lines contain descriptions of test cases — one per line. Each test case contains three integers $r$, $b$, $k$ ($1 \le r, b \le 10^9$, $2 \le k \le 10^9$) — the corresponding coefficients. -----Output----- Print $T$ words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise. -----Example----- Input 4 1 1 2 2 10 4 5 2 3 3 2 2 Output OBEY REBEL OBEY OBEY	def gcd(x, y): if x == 0: return y elif y == 0: return x elif x > y: return gcd(x % y, y) else: return gcd(x, y % x) t = int(input()) for i in range(t): r, b, k = map(int, input().split()) if r == b: print("obey") else: r, b = r / gcd(r, b), b / gcd(r, b) if b * (k - 1) <= r - 2 or r * (k - 1) <= b - 2: print("rebel") else: print("obey")	FUNC_DEF IF VAR NUMBER RETURN VAR IF VAR NUMBER RETURN VAR IF VAR VAR RETURN FUNC_CALL VAR BIN_OP VAR VAR VAR RETURN FUNC_CALL VAR VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR STRING ASSIGN VAR VAR BIN_OP VAR FUNC_CALL VAR VAR VAR BIN_OP VAR FUNC_CALL VAR VAR VAR IF BIN_OP VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER BIN_OP VAR BIN_OP VAR NUMBER BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor. You must paint a fence which consists of $10^{100}$ planks in two colors in the following way (suppose planks are numbered from left to right from $0$): if the index of the plank is divisible by $r$ (such planks have indices $0$, $r$, $2r$ and so on) then you must paint it red; if the index of the plank is divisible by $b$ (such planks have indices $0$, $b$, $2b$ and so on) then you must paint it blue; if the index is divisible both by $r$ and $b$ you can choose the color to paint the plank; otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it). Furthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are $k$ consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed. The question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs. -----Input----- The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. The next $T$ lines contain descriptions of test cases — one per line. Each test case contains three integers $r$, $b$, $k$ ($1 \le r, b \le 10^9$, $2 \le k \le 10^9$) — the corresponding coefficients. -----Output----- Print $T$ words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise. -----Example----- Input 4 1 1 2 2 10 4 5 2 3 3 2 2 Output OBEY REBEL OBEY OBEY	n = int(input()) def m(x, y): a = max(x, y) b = min(x, y) while a != b: c = a % b a = b b = c if c == 0: break return a for i in range(n): r, b, k = [int(x) for x in input().split(" ")] a = m(r, b) x = max(r, b) y = min(r, b) if 2 * x % y != 0: if x >= (k - 1) * y + 2 * a: print("REBEL") else: print("OBEY") elif x % y == 0: if x >= (k + 1) * y: print("REBEL") else: print("OBEY") elif x >= k * y + a: print("REBEL") else: print("OBEY")	ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR WHILE VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR IF VAR NUMBER RETURN VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF BIN_OP BIN_OP NUMBER VAR VAR NUMBER IF VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR BIN_OP NUMBER VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING IF BIN_OP VAR VAR NUMBER IF VAR BIN_OP BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING IF VAR BIN_OP BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor. You must paint a fence which consists of $10^{100}$ planks in two colors in the following way (suppose planks are numbered from left to right from $0$): if the index of the plank is divisible by $r$ (such planks have indices $0$, $r$, $2r$ and so on) then you must paint it red; if the index of the plank is divisible by $b$ (such planks have indices $0$, $b$, $2b$ and so on) then you must paint it blue; if the index is divisible both by $r$ and $b$ you can choose the color to paint the plank; otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it). Furthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are $k$ consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed. The question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs. -----Input----- The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. The next $T$ lines contain descriptions of test cases — one per line. Each test case contains three integers $r$, $b$, $k$ ($1 \le r, b \le 10^9$, $2 \le k \le 10^9$) — the corresponding coefficients. -----Output----- Print $T$ words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise. -----Example----- Input 4 1 1 2 2 10 4 5 2 3 3 2 2 Output OBEY REBEL OBEY OBEY	def gcd(a, b): while a % b != 0: r_ = a % b a = b b = r_ return b for _ in range(int(input())): r, b, k = map(int, input().split()) if r > b: r, b = b, r max_same = b // r g = gcd(r, b) r = r // g b = b // g if b % r == 0: max_same -= 1 if b % r > 1: max_same += 1 print("OBEY" if max_same < k else "REBEL")	FUNC_DEF WHILE BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR RETURN VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR IF BIN_OP VAR VAR NUMBER VAR NUMBER IF BIN_OP VAR VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR VAR STRING STRING
You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor. You must paint a fence which consists of $10^{100}$ planks in two colors in the following way (suppose planks are numbered from left to right from $0$): if the index of the plank is divisible by $r$ (such planks have indices $0$, $r$, $2r$ and so on) then you must paint it red; if the index of the plank is divisible by $b$ (such planks have indices $0$, $b$, $2b$ and so on) then you must paint it blue; if the index is divisible both by $r$ and $b$ you can choose the color to paint the plank; otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it). Furthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are $k$ consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed. The question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs. -----Input----- The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. The next $T$ lines contain descriptions of test cases — one per line. Each test case contains three integers $r$, $b$, $k$ ($1 \le r, b \le 10^9$, $2 \le k \le 10^9$) — the corresponding coefficients. -----Output----- Print $T$ words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise. -----Example----- Input 4 1 1 2 2 10 4 5 2 3 3 2 2 Output OBEY REBEL OBEY OBEY	def gcd(a, b): if b == 0: return a return gcd(b, a % b) def main(): a, b, k = map(int, input().split()) if a == b: print("OBEY") return l, m = min(a, b), max(a, b) d = gcd(a, b) diff = m - d c = int(diff / l) + 1 if diff % l == 0: c = c - 1 if c >= k: print("REBEL") else: print("OBEY") T = int(input()) for _ in range(T): main()	FUNC_DEF IF VAR NUMBER RETURN VAR RETURN FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_DEF ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR STRING RETURN ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR BIN_OP VAR VAR NUMBER IF BIN_OP VAR VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR
You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor. You must paint a fence which consists of $10^{100}$ planks in two colors in the following way (suppose planks are numbered from left to right from $0$): if the index of the plank is divisible by $r$ (such planks have indices $0$, $r$, $2r$ and so on) then you must paint it red; if the index of the plank is divisible by $b$ (such planks have indices $0$, $b$, $2b$ and so on) then you must paint it blue; if the index is divisible both by $r$ and $b$ you can choose the color to paint the plank; otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it). Furthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are $k$ consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed. The question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs. -----Input----- The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. The next $T$ lines contain descriptions of test cases — one per line. Each test case contains three integers $r$, $b$, $k$ ($1 \le r, b \le 10^9$, $2 \le k \le 10^9$) — the corresponding coefficients. -----Output----- Print $T$ words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise. -----Example----- Input 4 1 1 2 2 10 4 5 2 3 3 2 2 Output OBEY REBEL OBEY OBEY	import sys def gcd(a, b): if a == 0: return b return gcd(b % a, a) def solve(): import sys T = int(input()) while T > 0: rbk = sys.stdin.readline().split() r = int(rbk[0]) b = int(rbk[1]) k = int(rbk[2]) if r == b: print("OBEY") else: mx = max(r, b) mn = min(r, b) g = gcd(mx, mn) if g > 1: mx = mx / g mn = mn / g if 1 + (k - 1) * mn < mx: print("REBEL") else: print("OBEY") T -= 1 solve()	IMPORT FUNC_DEF IF VAR NUMBER RETURN VAR RETURN FUNC_CALL VAR BIN_OP VAR VAR VAR FUNC_DEF IMPORT ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR WHILE VAR NUMBER ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER ASSIGN VAR BIN_OP VAR VAR ASSIGN VAR BIN_OP VAR VAR IF BIN_OP NUMBER BIN_OP BIN_OP VAR NUMBER VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING VAR NUMBER EXPR FUNC_CALL VAR
You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor. You must paint a fence which consists of $10^{100}$ planks in two colors in the following way (suppose planks are numbered from left to right from $0$): if the index of the plank is divisible by $r$ (such planks have indices $0$, $r$, $2r$ and so on) then you must paint it red; if the index of the plank is divisible by $b$ (such planks have indices $0$, $b$, $2b$ and so on) then you must paint it blue; if the index is divisible both by $r$ and $b$ you can choose the color to paint the plank; otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it). Furthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are $k$ consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed. The question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs. -----Input----- The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. The next $T$ lines contain descriptions of test cases — one per line. Each test case contains three integers $r$, $b$, $k$ ($1 \le r, b \le 10^9$, $2 \le k \le 10^9$) — the corresponding coefficients. -----Output----- Print $T$ words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise. -----Example----- Input 4 1 1 2 2 10 4 5 2 3 3 2 2 Output OBEY REBEL OBEY OBEY	import sys input = iter(sys.stdin.read().splitlines()).__next__ def gcd(a, b): return a if b == 0 else gcd(b, a % b) def solveAH(r, b, k): if r == b: return True g = gcd(r, b) r //= g b //= g if r > b: r, b = b, r return (b - 1 + r - 1) // r < k def solveBF(r, b, k): if r == b: return True if r > b: r, b = b, r runR = 0 runB = 0 for n in range(10000): if n % r == 0 and n % b == 0: runR = 0 runB += 1 elif n % r == 0: runR += 1 runB = 0 elif n % b == 0: runB += 1 runR = 0 if runR >= k or runB >= k: return False return True TC = int(input()) for tc in range(TC): r, b, k = map(int, input().split()) if r > b: r, b = b, r res = solveAH(r, b, k) print("OBEY" if res else "REBEL")	IMPORT ASSIGN VAR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF RETURN VAR NUMBER VAR FUNC_CALL VAR VAR BIN_OP VAR VAR FUNC_DEF IF VAR VAR RETURN NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR IF VAR VAR ASSIGN VAR VAR VAR VAR RETURN BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER VAR NUMBER VAR VAR FUNC_DEF IF VAR VAR RETURN NUMBER IF VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER IF BIN_OP VAR VAR NUMBER BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER IF BIN_OP VAR VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER IF BIN_OP VAR VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER IF VAR VAR VAR VAR RETURN NUMBER RETURN NUMBER ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR VAR STRING STRING
You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor. You must paint a fence which consists of $10^{100}$ planks in two colors in the following way (suppose planks are numbered from left to right from $0$): if the index of the plank is divisible by $r$ (such planks have indices $0$, $r$, $2r$ and so on) then you must paint it red; if the index of the plank is divisible by $b$ (such planks have indices $0$, $b$, $2b$ and so on) then you must paint it blue; if the index is divisible both by $r$ and $b$ you can choose the color to paint the plank; otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it). Furthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are $k$ consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed. The question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs. -----Input----- The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. The next $T$ lines contain descriptions of test cases — one per line. Each test case contains three integers $r$, $b$, $k$ ($1 \le r, b \le 10^9$, $2 \le k \le 10^9$) — the corresponding coefficients. -----Output----- Print $T$ words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise. -----Example----- Input 4 1 1 2 2 10 4 5 2 3 3 2 2 Output OBEY REBEL OBEY OBEY	t = int(input()) def gcd(a, b): if b == 0: return a return gcd(b, a % b) for _ in range(t): r, b, k = map(int, input().split()) if r > b: r, b = b, r g = gcd(b, r) b /= g r /= g x = (b - 1 + r - 1) // r if x <= k - 1: print("OBEY") else: print("REBEL")	ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF IF VAR NUMBER RETURN VAR RETURN FUNC_CALL VAR VAR BIN_OP VAR VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR NUMBER VAR NUMBER VAR IF VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor. You must paint a fence which consists of $10^{100}$ planks in two colors in the following way (suppose planks are numbered from left to right from $0$): if the index of the plank is divisible by $r$ (such planks have indices $0$, $r$, $2r$ and so on) then you must paint it red; if the index of the plank is divisible by $b$ (such planks have indices $0$, $b$, $2b$ and so on) then you must paint it blue; if the index is divisible both by $r$ and $b$ you can choose the color to paint the plank; otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it). Furthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are $k$ consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed. The question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs. -----Input----- The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. The next $T$ lines contain descriptions of test cases — one per line. Each test case contains three integers $r$, $b$, $k$ ($1 \le r, b \le 10^9$, $2 \le k \le 10^9$) — the corresponding coefficients. -----Output----- Print $T$ words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise. -----Example----- Input 4 1 1 2 2 10 4 5 2 3 3 2 2 Output OBEY REBEL OBEY OBEY	import sys def gcd(x, y): while y != 0: x, y = y, x % y return x def main(): t = int(input()) allans = [] for _ in range(t): r, b, k = readIntArr() if r > b: temp = r r = b b = temp g = gcd(r, b) binn = 10*9 maxRebelK = 1 start = g while binn > 0: while start + r (maxRebelK - 1 + binn) < b: maxRebelK += binn binn //= 2 if k > maxRebelK: allans.append("OBEY") else: allans.append("REBEL") multiLineArrayPrint(allans) return input = lambda: sys.stdin.readline().rstrip("\r\n") def oneLineArrayPrint(arr): print(" ".join([str(x) for x in arr])) def multiLineArrayPrint(arr): print("\n".join([str(x) for x in arr])) def multiLineArrayOfArraysPrint(arr): print("\n".join([" ".join([str(x) for x in y]) for y in arr])) def readIntArr(): return [int(x) for x in input().split()] def makeArr(defaultVal, dimensionArr): dv = defaultVal da = dimensionArr if len(da) == 1: return [dv for _ in range(da[0])] else: return [makeArr(dv, da[1:]) for _ in range(da[0])] def queryInteractive(x, y): print("? {} {}".format(x, y)) sys.stdout.flush() return int(input()) def answerInteractive(ans): print("! {}".format(ans)) sys.stdout.flush() inf = float("inf") MOD = 10**9 + 7 for _abc in range(1): main()	IMPORT FUNC_DEF WHILE VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR VAR RETURN VAR FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR LIST FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR IF VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP NUMBER NUMBER ASSIGN VAR NUMBER ASSIGN VAR VAR WHILE VAR NUMBER WHILE BIN_OP VAR BIN_OP VAR BIN_OP BIN_OP VAR NUMBER VAR VAR VAR VAR VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR VAR RETURN ASSIGN VAR FUNC_CALL FUNC_CALL VAR STRING FUNC_DEF EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR VAR FUNC_DEF EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL VAR VAR VAR VAR FUNC_DEF EXPR FUNC_CALL VAR FUNC_CALL STRING FUNC_CALL STRING FUNC_CALL VAR VAR VAR VAR VAR VAR FUNC_DEF RETURN FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR FUNC_DEF ASSIGN VAR VAR ASSIGN VAR VAR IF FUNC_CALL VAR VAR NUMBER RETURN VAR VAR FUNC_CALL VAR VAR NUMBER RETURN FUNC_CALL VAR VAR VAR NUMBER VAR FUNC_CALL VAR VAR NUMBER FUNC_DEF EXPR FUNC_CALL VAR FUNC_CALL STRING VAR VAR EXPR FUNC_CALL VAR RETURN FUNC_CALL VAR FUNC_CALL VAR FUNC_DEF EXPR FUNC_CALL VAR FUNC_CALL STRING VAR EXPR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP BIN_OP NUMBER NUMBER NUMBER FOR VAR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR
You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor. You must paint a fence which consists of $10^{100}$ planks in two colors in the following way (suppose planks are numbered from left to right from $0$): if the index of the plank is divisible by $r$ (such planks have indices $0$, $r$, $2r$ and so on) then you must paint it red; if the index of the plank is divisible by $b$ (such planks have indices $0$, $b$, $2b$ and so on) then you must paint it blue; if the index is divisible both by $r$ and $b$ you can choose the color to paint the plank; otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it). Furthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are $k$ consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed. The question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs. -----Input----- The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. The next $T$ lines contain descriptions of test cases — one per line. Each test case contains three integers $r$, $b$, $k$ ($1 \le r, b \le 10^9$, $2 \le k \le 10^9$) — the corresponding coefficients. -----Output----- Print $T$ words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise. -----Example----- Input 4 1 1 2 2 10 4 5 2 3 3 2 2 Output OBEY REBEL OBEY OBEY	import sys readline = sys.stdin.readline readlines = sys.stdin.readlines ns = lambda: readline().rstrip() ni = lambda: int(readline().rstrip()) nm = lambda: map(int, readline().split()) nl = lambda: list(map(int, readline().split())) prn = lambda x: print(*x, sep="\n") def gcd(u, v): while v: u, v = v, u % v return u def solve(): r, b, k = nm() if r > b: r, b = b, r g = gcd(r, b) r //= g b //= g print("REBEL" if (b - 2) // r + 1 >= k else "OBEY") return T = ni() for _ in range(T): solve()	IMPORT ASSIGN VAR VAR ASSIGN VAR VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR STRING FUNC_DEF WHILE VAR ASSIGN VAR VAR VAR BIN_OP VAR VAR RETURN VAR FUNC_DEF ASSIGN VAR VAR VAR FUNC_CALL VAR IF VAR VAR ASSIGN VAR VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP BIN_OP BIN_OP VAR NUMBER VAR NUMBER VAR STRING STRING RETURN ASSIGN VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR
You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor. You must paint a fence which consists of $10^{100}$ planks in two colors in the following way (suppose planks are numbered from left to right from $0$): if the index of the plank is divisible by $r$ (such planks have indices $0$, $r$, $2r$ and so on) then you must paint it red; if the index of the plank is divisible by $b$ (such planks have indices $0$, $b$, $2b$ and so on) then you must paint it blue; if the index is divisible both by $r$ and $b$ you can choose the color to paint the plank; otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it). Furthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are $k$ consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed. The question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs. -----Input----- The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. The next $T$ lines contain descriptions of test cases — one per line. Each test case contains three integers $r$, $b$, $k$ ($1 \le r, b \le 10^9$, $2 \le k \le 10^9$) — the corresponding coefficients. -----Output----- Print $T$ words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise. -----Example----- Input 4 1 1 2 2 10 4 5 2 3 3 2 2 Output OBEY REBEL OBEY OBEY	t = int(input()) for _ in range(t): r, b, k = map(int, input().split(" ")) if r > b: r, b = b, r if b % r == 0: if b // r - 1 >= k: print("REBEL") else: print("OBEY") continue ok = True diff = r - ((b // r + 1) * r - b) x = r - r // diff * diff if r % diff == 0: x += diff if (b - x) // r + ((b - x) % r != 0) >= k: ok = False if ok: print("OBEY") else: print("REBEL")	ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR STRING IF VAR VAR ASSIGN VAR VAR VAR VAR IF BIN_OP VAR VAR NUMBER IF BIN_OP BIN_OP VAR VAR NUMBER VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR NUMBER VAR VAR ASSIGN VAR BIN_OP VAR BIN_OP BIN_OP VAR VAR VAR IF BIN_OP VAR VAR NUMBER VAR VAR IF BIN_OP BIN_OP BIN_OP VAR VAR VAR BIN_OP BIN_OP VAR VAR VAR NUMBER VAR ASSIGN VAR NUMBER IF VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor. You must paint a fence which consists of $10^{100}$ planks in two colors in the following way (suppose planks are numbered from left to right from $0$): if the index of the plank is divisible by $r$ (such planks have indices $0$, $r$, $2r$ and so on) then you must paint it red; if the index of the plank is divisible by $b$ (such planks have indices $0$, $b$, $2b$ and so on) then you must paint it blue; if the index is divisible both by $r$ and $b$ you can choose the color to paint the plank; otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it). Furthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are $k$ consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed. The question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs. -----Input----- The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. The next $T$ lines contain descriptions of test cases — one per line. Each test case contains three integers $r$, $b$, $k$ ($1 \le r, b \le 10^9$, $2 \le k \le 10^9$) — the corresponding coefficients. -----Output----- Print $T$ words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise. -----Example----- Input 4 1 1 2 2 10 4 5 2 3 3 2 2 Output OBEY REBEL OBEY OBEY	def gcd(a, b): while b: a, b = b, a % b return a T = int(input()) for _ in range(T): r, b, k = map(int, input().split()) r, b = min(r, b), max(r, b) if gcd(r, b) + r * (k - 1) < b: print("REBEL") else: print("OBEY")	FUNC_DEF WHILE VAR ASSIGN VAR VAR VAR BIN_OP VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF BIN_OP FUNC_CALL VAR VAR VAR BIN_OP VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor. You must paint a fence which consists of $10^{100}$ planks in two colors in the following way (suppose planks are numbered from left to right from $0$): if the index of the plank is divisible by $r$ (such planks have indices $0$, $r$, $2r$ and so on) then you must paint it red; if the index of the plank is divisible by $b$ (such planks have indices $0$, $b$, $2b$ and so on) then you must paint it blue; if the index is divisible both by $r$ and $b$ you can choose the color to paint the plank; otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it). Furthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are $k$ consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed. The question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs. -----Input----- The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. The next $T$ lines contain descriptions of test cases — one per line. Each test case contains three integers $r$, $b$, $k$ ($1 \le r, b \le 10^9$, $2 \le k \le 10^9$) — the corresponding coefficients. -----Output----- Print $T$ words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise. -----Example----- Input 4 1 1 2 2 10 4 5 2 3 3 2 2 Output OBEY REBEL OBEY OBEY	def nod(a, b): a, b = min(a, b), max(a, b) while a > 0: a, b = b % a, a return b n = int(input()) for i in range(n): r, b, k = [int(i) for i in input().split()] r, b = max(r, b), min(r, b) if (k - 1) * b + nod(b, r) >= r: print("OBEY") else: print("REBEL")	FUNC_DEF ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR WHILE VAR NUMBER ASSIGN VAR VAR BIN_OP VAR VAR VAR RETURN VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF BIN_OP BIN_OP BIN_OP VAR NUMBER VAR FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor. You must paint a fence which consists of $10^{100}$ planks in two colors in the following way (suppose planks are numbered from left to right from $0$): if the index of the plank is divisible by $r$ (such planks have indices $0$, $r$, $2r$ and so on) then you must paint it red; if the index of the plank is divisible by $b$ (such planks have indices $0$, $b$, $2b$ and so on) then you must paint it blue; if the index is divisible both by $r$ and $b$ you can choose the color to paint the plank; otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it). Furthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are $k$ consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed. The question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs. -----Input----- The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. The next $T$ lines contain descriptions of test cases — one per line. Each test case contains three integers $r$, $b$, $k$ ($1 \le r, b \le 10^9$, $2 \le k \le 10^9$) — the corresponding coefficients. -----Output----- Print $T$ words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise. -----Example----- Input 4 1 1 2 2 10 4 5 2 3 3 2 2 Output OBEY REBEL OBEY OBEY	for t in range(int(input())): r, b, k = map(int, input().split()) if b > r: b, r = r, b if r == b: print("OBEY") elif r % b == 0: if r // b <= k: print("OBEY") else: print("REBEL") elif r % (r % b) == 0 and b % (r % b) == 0: if r // b < k: print("OBEY") else: print("REBEL") elif r // b + 1 < k: print("OBEY") else: print("REBEL")	FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR ASSIGN VAR VAR VAR VAR IF VAR VAR EXPR FUNC_CALL VAR STRING IF BIN_OP VAR VAR NUMBER IF BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING IF BIN_OP VAR BIN_OP VAR VAR NUMBER BIN_OP VAR BIN_OP VAR VAR NUMBER IF BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING IF BIN_OP BIN_OP VAR VAR NUMBER VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
You are a rebel leader and you are planning to start a revolution in your country. But the evil Government found out about your plans and set your punishment in the form of correctional labor. You must paint a fence which consists of $10^{100}$ planks in two colors in the following way (suppose planks are numbered from left to right from $0$): if the index of the plank is divisible by $r$ (such planks have indices $0$, $r$, $2r$ and so on) then you must paint it red; if the index of the plank is divisible by $b$ (such planks have indices $0$, $b$, $2b$ and so on) then you must paint it blue; if the index is divisible both by $r$ and $b$ you can choose the color to paint the plank; otherwise, you don't need to paint the plank at all (and it is forbidden to spent paint on it). Furthermore, the Government added one additional restriction to make your punishment worse. Let's list all painted planks of the fence in ascending order: if there are $k$ consecutive planks with the same color in this list, then the Government will state that you failed the labor and execute you immediately. If you don't paint the fence according to the four aforementioned conditions, you will also be executed. The question is: will you be able to accomplish the labor (the time is not important) or the execution is unavoidable and you need to escape at all costs. -----Input----- The first line contains single integer $T$ ($1 \le T \le 1000$) — the number of test cases. The next $T$ lines contain descriptions of test cases — one per line. Each test case contains three integers $r$, $b$, $k$ ($1 \le r, b \le 10^9$, $2 \le k \le 10^9$) — the corresponding coefficients. -----Output----- Print $T$ words — one per line. For each test case print REBEL (case insensitive) if the execution is unavoidable or OBEY (case insensitive) otherwise. -----Example----- Input 4 1 1 2 2 10 4 5 2 3 3 2 2 Output OBEY REBEL OBEY OBEY	def euc(a, b): while b != 0: a, b = b, a % b return a for _ in range(int(input())): r, b, k = map(int, input().split()) if r == b: print("OBEY") continue r, b = min(r, b), max(r, b) if b % r == 0: if b // r - 1 < k: print("OBEY") else: print("REBEL") continue a = euc(b, r) b //= a r //= a if r * k - (r - 1) >= b: print("OBEY") continue print("REBEL")	FUNC_DEF WHILE VAR NUMBER ASSIGN VAR VAR VAR BIN_OP VAR VAR RETURN VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR IF VAR VAR EXPR FUNC_CALL VAR STRING ASSIGN VAR VAR FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR VAR IF BIN_OP VAR VAR NUMBER IF BIN_OP BIN_OP VAR VAR NUMBER VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR VAR VAR VAR VAR VAR IF BIN_OP BIN_OP VAR VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING
Little Petya very much likes playing with little Masha. Recently he has received a game called "Zero-One" as a gift from his mother. Petya immediately offered Masha to play the game with him. Before the very beginning of the game several cards are lain out on a table in one line from the left to the right. Each card contains a digit: 0 or 1. Players move in turns and Masha moves first. During each move a player should remove a card from the table and shift all other cards so as to close the gap left by the removed card. For example, if before somebody's move the cards on the table formed a sequence 01010101, then after the fourth card is removed (the cards are numbered starting from 1), the sequence will look like that: 0100101. The game ends when exactly two cards are left on the table. The digits on these cards determine the number in binary notation: the most significant bit is located to the left. Masha's aim is to minimize the number and Petya's aim is to maximize it. An unpleasant accident occurred before the game started. The kids spilled juice on some of the cards and the digits on the cards got blurred. Each one of the spoiled cards could have either 0 or 1 written on it. Consider all possible variants of initial arrangement of the digits (before the juice spilling). For each variant, let's find which two cards are left by the end of the game, assuming that both Petya and Masha play optimally. An ordered pair of digits written on those two cards is called an outcome. Your task is to find the set of outcomes for all variants of initial digits arrangement. Input The first line contains a sequence of characters each of which can either be a "0", a "1" or a "?". This sequence determines the initial arrangement of cards on the table from the left to the right. The characters "?" mean that the given card was spoiled before the game. The sequence's length ranges from 2 to 105, inclusive. Output Print the set of outcomes for all possible initial digits arrangements. Print each possible outcome on a single line. Each outcome should be represented by two characters: the digits written on the cards that were left by the end of the game. The outcomes should be sorted lexicographically in ascending order (see the first sample). Examples Input ???? Output 00 01 10 11 Input 1010 Output 10 Input 1?1 Output 01 11 Note In the first sample all 16 variants of numbers arrangement are possible. For the variant 0000 the outcome is 00. For the variant 1111 the outcome is 11. For the variant 0011 the outcome is 01. For the variant 1100 the outcome is 10. Regardless of outcomes for all other variants the set which we are looking for will contain all 4 possible outcomes. In the third sample only 2 variants of numbers arrangement are possible: 111 and 101. For the variant 111 the outcome is 11. For the variant 101 the outcome is 01, because on the first turn Masha can remove the first card from the left after which the game will end.	s = input() a, b, c = (s.count(x) for x in "01?") if a + c > b: print("00") if a + c + 2 > b >= a - c: if s[-1] != "?": if s[-1] == "0": print("10") else: print("01") else: if a + c > b: print("01") if b + c > a + 1: print("10") if b + c > a + 1: print("11")	ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR VAR FUNC_CALL VAR VAR VAR STRING IF BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR STRING IF BIN_OP BIN_OP VAR VAR NUMBER VAR BIN_OP VAR VAR IF VAR NUMBER STRING IF VAR NUMBER STRING EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR STRING IF BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR STRING IF BIN_OP VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING IF BIN_OP VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING
Little Petya very much likes playing with little Masha. Recently he has received a game called "Zero-One" as a gift from his mother. Petya immediately offered Masha to play the game with him. Before the very beginning of the game several cards are lain out on a table in one line from the left to the right. Each card contains a digit: 0 or 1. Players move in turns and Masha moves first. During each move a player should remove a card from the table and shift all other cards so as to close the gap left by the removed card. For example, if before somebody's move the cards on the table formed a sequence 01010101, then after the fourth card is removed (the cards are numbered starting from 1), the sequence will look like that: 0100101. The game ends when exactly two cards are left on the table. The digits on these cards determine the number in binary notation: the most significant bit is located to the left. Masha's aim is to minimize the number and Petya's aim is to maximize it. An unpleasant accident occurred before the game started. The kids spilled juice on some of the cards and the digits on the cards got blurred. Each one of the spoiled cards could have either 0 or 1 written on it. Consider all possible variants of initial arrangement of the digits (before the juice spilling). For each variant, let's find which two cards are left by the end of the game, assuming that both Petya and Masha play optimally. An ordered pair of digits written on those two cards is called an outcome. Your task is to find the set of outcomes for all variants of initial digits arrangement. Input The first line contains a sequence of characters each of which can either be a "0", a "1" or a "?". This sequence determines the initial arrangement of cards on the table from the left to the right. The characters "?" mean that the given card was spoiled before the game. The sequence's length ranges from 2 to 105, inclusive. Output Print the set of outcomes for all possible initial digits arrangements. Print each possible outcome on a single line. Each outcome should be represented by two characters: the digits written on the cards that were left by the end of the game. The outcomes should be sorted lexicographically in ascending order (see the first sample). Examples Input ???? Output 00 01 10 11 Input 1010 Output 10 Input 1?1 Output 01 11 Note In the first sample all 16 variants of numbers arrangement are possible. For the variant 0000 the outcome is 00. For the variant 1111 the outcome is 11. For the variant 0011 the outcome is 01. For the variant 1100 the outcome is 10. Regardless of outcomes for all other variants the set which we are looking for will contain all 4 possible outcomes. In the third sample only 2 variants of numbers arrangement are possible: 111 and 101. For the variant 111 the outcome is 11. For the variant 101 the outcome is 01, because on the first turn Masha can remove the first card from the left after which the game will end.	s = input() n = len(s) zeros, ones, other = 0, 0, 0 for x in s: zeros += x == "0" ones += x == "1" other += x == "?" A = ["00", "01", "10", "11"] pos = [0] * 4 pos[0] = zeros + other > n // 2 pos[3] = zeros < n // 2 if zeros <= n // 2 and zeros + other >= n // 2: canuse = other - n // 2 + zeros pos[1] = s[n - 1] == "1" or s[n - 1] == "?" and canuse canuse = zeros < n // 2 pos[2] = s[n - 1] == "0" or s[n - 1] == "?" and canuse for i in range(4): if pos[i]: print(A[i])	ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR VAR VAR NUMBER NUMBER NUMBER FOR VAR VAR VAR VAR STRING VAR VAR STRING VAR VAR STRING ASSIGN VAR LIST STRING STRING STRING STRING ASSIGN VAR BIN_OP LIST NUMBER NUMBER ASSIGN VAR NUMBER BIN_OP VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER VAR BIN_OP VAR NUMBER IF VAR BIN_OP VAR NUMBER BIN_OP VAR VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR BIN_OP VAR NUMBER VAR ASSIGN VAR NUMBER VAR BIN_OP VAR NUMBER STRING VAR BIN_OP VAR NUMBER STRING VAR ASSIGN VAR VAR BIN_OP VAR NUMBER ASSIGN VAR NUMBER VAR BIN_OP VAR NUMBER STRING VAR BIN_OP VAR NUMBER STRING VAR FOR VAR FUNC_CALL VAR NUMBER IF VAR VAR EXPR FUNC_CALL VAR VAR VAR
Little Petya very much likes playing with little Masha. Recently he has received a game called "Zero-One" as a gift from his mother. Petya immediately offered Masha to play the game with him. Before the very beginning of the game several cards are lain out on a table in one line from the left to the right. Each card contains a digit: 0 or 1. Players move in turns and Masha moves first. During each move a player should remove a card from the table and shift all other cards so as to close the gap left by the removed card. For example, if before somebody's move the cards on the table formed a sequence 01010101, then after the fourth card is removed (the cards are numbered starting from 1), the sequence will look like that: 0100101. The game ends when exactly two cards are left on the table. The digits on these cards determine the number in binary notation: the most significant bit is located to the left. Masha's aim is to minimize the number and Petya's aim is to maximize it. An unpleasant accident occurred before the game started. The kids spilled juice on some of the cards and the digits on the cards got blurred. Each one of the spoiled cards could have either 0 or 1 written on it. Consider all possible variants of initial arrangement of the digits (before the juice spilling). For each variant, let's find which two cards are left by the end of the game, assuming that both Petya and Masha play optimally. An ordered pair of digits written on those two cards is called an outcome. Your task is to find the set of outcomes for all variants of initial digits arrangement. Input The first line contains a sequence of characters each of which can either be a "0", a "1" or a "?". This sequence determines the initial arrangement of cards on the table from the left to the right. The characters "?" mean that the given card was spoiled before the game. The sequence's length ranges from 2 to 105, inclusive. Output Print the set of outcomes for all possible initial digits arrangements. Print each possible outcome on a single line. Each outcome should be represented by two characters: the digits written on the cards that were left by the end of the game. The outcomes should be sorted lexicographically in ascending order (see the first sample). Examples Input ???? Output 00 01 10 11 Input 1010 Output 10 Input 1?1 Output 01 11 Note In the first sample all 16 variants of numbers arrangement are possible. For the variant 0000 the outcome is 00. For the variant 1111 the outcome is 11. For the variant 0011 the outcome is 01. For the variant 1100 the outcome is 10. Regardless of outcomes for all other variants the set which we are looking for will contain all 4 possible outcomes. In the third sample only 2 variants of numbers arrangement are possible: 111 and 101. For the variant 111 the outcome is 11. For the variant 101 the outcome is 01, because on the first turn Masha can remove the first card from the left after which the game will end.	res = input() possible = [] zero = res.count("0") one = res.count("1") question = res.count("?") if (len(res) - 2) // 2 + 2 <= zero + question: possible.append("00") if res[-1] == "?": if ( one + 1 <= (len(res) + 1) // 2 <= one + question and zero <= len(res) // 2 <= zero + question - 1 ): possible.append("01") if ( one <= (len(res) + 1) // 2 <= one + question - 1 and zero + 1 <= len(res) // 2 <= zero + question ): possible.append("10") elif ( one <= (len(res) + 1) // 2 <= one + question and zero <= len(res) // 2 <= zero + question ): if res[-1] == "1": possible.append("01") if res[-1] == "0": possible.append("10") if (len(res) - 1) // 2 + 2 <= one + question: possible.append("11") print("\n".join(possible))	ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING IF BIN_OP BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER BIN_OP VAR VAR EXPR FUNC_CALL VAR STRING IF VAR NUMBER STRING IF BIN_OP VAR NUMBER BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER BIN_OP VAR VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR STRING IF VAR BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER BIN_OP BIN_OP VAR VAR NUMBER BIN_OP VAR NUMBER BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP VAR VAR EXPR FUNC_CALL VAR STRING IF VAR BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER BIN_OP VAR VAR VAR BIN_OP FUNC_CALL VAR VAR NUMBER BIN_OP VAR VAR IF VAR NUMBER STRING EXPR FUNC_CALL VAR STRING IF VAR NUMBER STRING EXPR FUNC_CALL VAR STRING IF BIN_OP BIN_OP BIN_OP FUNC_CALL VAR VAR NUMBER NUMBER NUMBER BIN_OP VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR FUNC_CALL STRING VAR
Little Petya very much likes playing with little Masha. Recently he has received a game called "Zero-One" as a gift from his mother. Petya immediately offered Masha to play the game with him. Before the very beginning of the game several cards are lain out on a table in one line from the left to the right. Each card contains a digit: 0 or 1. Players move in turns and Masha moves first. During each move a player should remove a card from the table and shift all other cards so as to close the gap left by the removed card. For example, if before somebody's move the cards on the table formed a sequence 01010101, then after the fourth card is removed (the cards are numbered starting from 1), the sequence will look like that: 0100101. The game ends when exactly two cards are left on the table. The digits on these cards determine the number in binary notation: the most significant bit is located to the left. Masha's aim is to minimize the number and Petya's aim is to maximize it. An unpleasant accident occurred before the game started. The kids spilled juice on some of the cards and the digits on the cards got blurred. Each one of the spoiled cards could have either 0 or 1 written on it. Consider all possible variants of initial arrangement of the digits (before the juice spilling). For each variant, let's find which two cards are left by the end of the game, assuming that both Petya and Masha play optimally. An ordered pair of digits written on those two cards is called an outcome. Your task is to find the set of outcomes for all variants of initial digits arrangement. Input The first line contains a sequence of characters each of which can either be a "0", a "1" or a "?". This sequence determines the initial arrangement of cards on the table from the left to the right. The characters "?" mean that the given card was spoiled before the game. The sequence's length ranges from 2 to 105, inclusive. Output Print the set of outcomes for all possible initial digits arrangements. Print each possible outcome on a single line. Each outcome should be represented by two characters: the digits written on the cards that were left by the end of the game. The outcomes should be sorted lexicographically in ascending order (see the first sample). Examples Input ???? Output 00 01 10 11 Input 1010 Output 10 Input 1?1 Output 01 11 Note In the first sample all 16 variants of numbers arrangement are possible. For the variant 0000 the outcome is 00. For the variant 1111 the outcome is 11. For the variant 0011 the outcome is 01. For the variant 1100 the outcome is 10. Regardless of outcomes for all other variants the set which we are looking for will contain all 4 possible outcomes. In the third sample only 2 variants of numbers arrangement are possible: 111 and 101. For the variant 111 the outcome is 11. For the variant 101 the outcome is 01, because on the first turn Masha can remove the first card from the left after which the game will end.	def evaluate(a): c1 = a.count("1") c0 = a.count("0") n = len(a) A = (n - 1) // 2 B = (n - 2) // 2 if c1 <= A: return "00" if c0 <= B: return "11" p1 = a.rfind("1") p0 = a.rfind("0") if p0 < p1: return "01" else: return "10" a = input() x = [] x.append(evaluate(a.replace("?", "0"))) x.append(evaluate(a.replace("?", "1"))) n = len(a) c1 = a.count("1") c0 = a.count("0") A = (n - 1) // 2 B = (n - 2) // 2 x.append(evaluate(a.replace("?", "0", B + 1 - c0).replace("?", "1"))) x.append(evaluate(a.replace("?", "1", A + 1 - c1).replace("?", "0"))) for ans in sorted(list(set(x))): print(ans)	FUNC_DEF ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER IF VAR VAR RETURN STRING IF VAR VAR RETURN STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING IF VAR VAR RETURN STRING RETURN STRING ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR STRING STRING EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR STRING STRING ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR STRING STRING BIN_OP BIN_OP VAR NUMBER VAR STRING STRING EXPR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR STRING STRING BIN_OP BIN_OP VAR NUMBER VAR STRING STRING FOR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR VAR
Little Petya very much likes playing with little Masha. Recently he has received a game called "Zero-One" as a gift from his mother. Petya immediately offered Masha to play the game with him. Before the very beginning of the game several cards are lain out on a table in one line from the left to the right. Each card contains a digit: 0 or 1. Players move in turns and Masha moves first. During each move a player should remove a card from the table and shift all other cards so as to close the gap left by the removed card. For example, if before somebody's move the cards on the table formed a sequence 01010101, then after the fourth card is removed (the cards are numbered starting from 1), the sequence will look like that: 0100101. The game ends when exactly two cards are left on the table. The digits on these cards determine the number in binary notation: the most significant bit is located to the left. Masha's aim is to minimize the number and Petya's aim is to maximize it. An unpleasant accident occurred before the game started. The kids spilled juice on some of the cards and the digits on the cards got blurred. Each one of the spoiled cards could have either 0 or 1 written on it. Consider all possible variants of initial arrangement of the digits (before the juice spilling). For each variant, let's find which two cards are left by the end of the game, assuming that both Petya and Masha play optimally. An ordered pair of digits written on those two cards is called an outcome. Your task is to find the set of outcomes for all variants of initial digits arrangement. Input The first line contains a sequence of characters each of which can either be a "0", a "1" or a "?". This sequence determines the initial arrangement of cards on the table from the left to the right. The characters "?" mean that the given card was spoiled before the game. The sequence's length ranges from 2 to 105, inclusive. Output Print the set of outcomes for all possible initial digits arrangements. Print each possible outcome on a single line. Each outcome should be represented by two characters: the digits written on the cards that were left by the end of the game. The outcomes should be sorted lexicographically in ascending order (see the first sample). Examples Input ???? Output 00 01 10 11 Input 1010 Output 10 Input 1?1 Output 01 11 Note In the first sample all 16 variants of numbers arrangement are possible. For the variant 0000 the outcome is 00. For the variant 1111 the outcome is 11. For the variant 0011 the outcome is 01. For the variant 1100 the outcome is 10. Regardless of outcomes for all other variants the set which we are looking for will contain all 4 possible outcomes. In the third sample only 2 variants of numbers arrangement are possible: 111 and 101. For the variant 111 the outcome is 11. For the variant 101 the outcome is 01, because on the first turn Masha can remove the first card from the left after which the game will end.	import sys def rint(): return map(int, sys.stdin.readline().split()) s = list(input()) q = s.count("?") o = s.count("1") z = s.count("0") ans = [] if z + q >= o + 1: ans.append("00") tmp = o + q - z - (z + o + q) % 2 if tmp % 2 == 0: x = tmp // 2 if x >= 0 and x <= q: if s[-1] == "1" or s[-1] == "?" and q - x > 0: ans.append("01") tmp = z + q - o + (z + o + q) % 2 if tmp % 2 == 0: x = tmp // 2 if x >= 0 and x <= q: if s[-1] == "0" or s[-1] == "?" and q - x > 0: ans.append("10") if o + q >= z + 2: ans.append("11") for a in ans: print(a)	IMPORT FUNC_DEF RETURN FUNC_CALL VAR VAR FUNC_CALL FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR STRING ASSIGN VAR LIST IF BIN_OP VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF VAR NUMBER VAR VAR IF VAR NUMBER STRING VAR NUMBER STRING BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR STRING ASSIGN VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER IF BIN_OP VAR NUMBER NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF VAR NUMBER VAR VAR IF VAR NUMBER STRING VAR NUMBER STRING BIN_OP VAR VAR NUMBER EXPR FUNC_CALL VAR STRING IF BIN_OP VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING FOR VAR VAR EXPR FUNC_CALL VAR VAR
Little Petya very much likes playing with little Masha. Recently he has received a game called "Zero-One" as a gift from his mother. Petya immediately offered Masha to play the game with him. Before the very beginning of the game several cards are lain out on a table in one line from the left to the right. Each card contains a digit: 0 or 1. Players move in turns and Masha moves first. During each move a player should remove a card from the table and shift all other cards so as to close the gap left by the removed card. For example, if before somebody's move the cards on the table formed a sequence 01010101, then after the fourth card is removed (the cards are numbered starting from 1), the sequence will look like that: 0100101. The game ends when exactly two cards are left on the table. The digits on these cards determine the number in binary notation: the most significant bit is located to the left. Masha's aim is to minimize the number and Petya's aim is to maximize it. An unpleasant accident occurred before the game started. The kids spilled juice on some of the cards and the digits on the cards got blurred. Each one of the spoiled cards could have either 0 or 1 written on it. Consider all possible variants of initial arrangement of the digits (before the juice spilling). For each variant, let's find which two cards are left by the end of the game, assuming that both Petya and Masha play optimally. An ordered pair of digits written on those two cards is called an outcome. Your task is to find the set of outcomes for all variants of initial digits arrangement. Input The first line contains a sequence of characters each of which can either be a "0", a "1" or a "?". This sequence determines the initial arrangement of cards on the table from the left to the right. The characters "?" mean that the given card was spoiled before the game. The sequence's length ranges from 2 to 105, inclusive. Output Print the set of outcomes for all possible initial digits arrangements. Print each possible outcome on a single line. Each outcome should be represented by two characters: the digits written on the cards that were left by the end of the game. The outcomes should be sorted lexicographically in ascending order (see the first sample). Examples Input ???? Output 00 01 10 11 Input 1010 Output 10 Input 1?1 Output 01 11 Note In the first sample all 16 variants of numbers arrangement are possible. For the variant 0000 the outcome is 00. For the variant 1111 the outcome is 11. For the variant 0011 the outcome is 01. For the variant 1100 the outcome is 10. Regardless of outcomes for all other variants the set which we are looking for will contain all 4 possible outcomes. In the third sample only 2 variants of numbers arrangement are possible: 111 and 101. For the variant 111 the outcome is 11. For the variant 101 the outcome is 01, because on the first turn Masha can remove the first card from the left after which the game will end.	a, b, c = 0, 0, 0 last = "" for i in input(): if i == "1": a += 1 elif i == "0": b += 1 else: c += 1 last = i if b + c > a: print("00") if last != "0": a1, b1, c1 = a, b, c if last == "?": c1 -= 1 a1 += 1 x = (-a1 + b1 + c1 + (a1 + b1 + c1) % 2) // 2 if 0 <= x <= c1: print("01") if last != "1": a1, b1, c1 = a, b, c if last == "?": c1 -= 1 b1 += 1 x = (-a1 + b1 + c1 + (a1 + b1 + c1) % 2) // 2 if 0 <= x <= c1: print("10") if a + c > b + 1: print("11")	ASSIGN VAR VAR VAR NUMBER NUMBER NUMBER ASSIGN VAR STRING FOR VAR FUNC_CALL VAR IF VAR STRING VAR NUMBER IF VAR STRING VAR NUMBER VAR NUMBER ASSIGN VAR VAR IF BIN_OP VAR VAR VAR EXPR FUNC_CALL VAR STRING IF VAR STRING ASSIGN VAR VAR VAR VAR VAR VAR IF VAR STRING VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER NUMBER IF NUMBER VAR VAR EXPR FUNC_CALL VAR STRING IF VAR STRING ASSIGN VAR VAR VAR VAR VAR VAR IF VAR STRING VAR NUMBER VAR NUMBER ASSIGN VAR BIN_OP BIN_OP BIN_OP BIN_OP VAR VAR VAR BIN_OP BIN_OP BIN_OP VAR VAR VAR NUMBER NUMBER IF NUMBER VAR VAR EXPR FUNC_CALL VAR STRING IF BIN_OP VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR STRING
Fangy the little walrus, as all the modern walruses, loves to communicate via text messaging. One day he faced the following problem: When he sends large texts, they are split into parts each containing n characters (which is the size of one text message). Thus, whole sentences and words get split! Fangy did not like it, so he faced the task of breaking the text into minimal messages on his own so that no sentence were broken into pieces when it is sent and the number of text messages to be sent would be minimal. If two consecutive sentences are in different messages, the space between them can be ignored (Fangy does not write this space). The little walrus's text looks in the following manner: TEXT ::= SENTENCE \| SENTENCE SPACE TEXT SENTENCE ::= WORD SPACE SENTENCE \| WORD END END ::= {'.', '?', '!'} WORD ::= LETTER \| LETTER WORD LETTER ::= {'a'..'z', 'A'..'Z'} SPACE ::= ' ' SPACE stands for the symbol of a space. So, how many messages did Fangy send? Input The first line contains an integer n, which is the size of one message (2 ≤ n ≤ 255). The second line contains the text. The length of the text does not exceed 104 characters. It is guaranteed that the text satisfies the above described format. Specifically, this implies that the text is not empty. Output On the first and only line print the number of text messages Fangy will need. If it is impossible to split the text, print "Impossible" without the quotes. Examples Input 25 Hello. I am a little walrus. Output 2 Input 2 How are you? Output Impossible Input 19 Hello! Do you like fish? Why? Output 3 Note Let's take a look at the third sample. The text will be split into three messages: "Hello!", "Do you like fish?" and "Why?".	n, text, SMSes, SMS_len = int(input()), input(), 0, 0 for ch in ".?!": text = text.replace(ch, "_") for L in map(len, (" " + text).split("_")[:-1]): if L > n: print("Impossible") break SMS_len += L + 1 if SMS_len else L if SMS_len > n: SMSes, SMS_len = SMSes + 1, L else: print(SMSes + 1)	ASSIGN VAR VAR VAR VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR NUMBER NUMBER FOR VAR STRING ASSIGN VAR FUNC_CALL VAR VAR STRING FOR VAR FUNC_CALL VAR VAR FUNC_CALL BIN_OP STRING VAR STRING NUMBER IF VAR VAR EXPR FUNC_CALL VAR STRING VAR VAR BIN_OP VAR NUMBER VAR IF VAR VAR ASSIGN VAR VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER
Fangy the little walrus, as all the modern walruses, loves to communicate via text messaging. One day he faced the following problem: When he sends large texts, they are split into parts each containing n characters (which is the size of one text message). Thus, whole sentences and words get split! Fangy did not like it, so he faced the task of breaking the text into minimal messages on his own so that no sentence were broken into pieces when it is sent and the number of text messages to be sent would be minimal. If two consecutive sentences are in different messages, the space between them can be ignored (Fangy does not write this space). The little walrus's text looks in the following manner: TEXT ::= SENTENCE \| SENTENCE SPACE TEXT SENTENCE ::= WORD SPACE SENTENCE \| WORD END END ::= {'.', '?', '!'} WORD ::= LETTER \| LETTER WORD LETTER ::= {'a'..'z', 'A'..'Z'} SPACE ::= ' ' SPACE stands for the symbol of a space. So, how many messages did Fangy send? Input The first line contains an integer n, which is the size of one message (2 ≤ n ≤ 255). The second line contains the text. The length of the text does not exceed 104 characters. It is guaranteed that the text satisfies the above described format. Specifically, this implies that the text is not empty. Output On the first and only line print the number of text messages Fangy will need. If it is impossible to split the text, print "Impossible" without the quotes. Examples Input 25 Hello. I am a little walrus. Output 2 Input 2 How are you? Output Impossible Input 19 Hello! Do you like fish? Why? Output 3 Note Let's take a look at the third sample. The text will be split into three messages: "Hello!", "Do you like fish?" and "Why?".	import sys n = int(input()) s = str(input()) m = len(s) cnt = 0 gd = False ans = 0 lst = 0 end = [".", "?", "!"] rem = 0 for i in range(m): cnt += 1 if s[i] in end: gd = True lst = cnt if cnt == n: if not gd: print("Impossible") exit(0) else: cnt = cnt - lst - 1 gd = False ans += 1 rem = i if s[m - 1] not in end: print("Impossible") exit(0) elif rem != m - 1: ans += 1 print(ans)	IMPORT ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR LIST STRING STRING STRING ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR VAR VAR NUMBER IF VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR VAR IF VAR VAR IF VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER ASSIGN VAR BIN_OP BIN_OP VAR VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER ASSIGN VAR VAR IF VAR BIN_OP VAR NUMBER VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER IF VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR
Fangy the little walrus, as all the modern walruses, loves to communicate via text messaging. One day he faced the following problem: When he sends large texts, they are split into parts each containing n characters (which is the size of one text message). Thus, whole sentences and words get split! Fangy did not like it, so he faced the task of breaking the text into minimal messages on his own so that no sentence were broken into pieces when it is sent and the number of text messages to be sent would be minimal. If two consecutive sentences are in different messages, the space between them can be ignored (Fangy does not write this space). The little walrus's text looks in the following manner: TEXT ::= SENTENCE \| SENTENCE SPACE TEXT SENTENCE ::= WORD SPACE SENTENCE \| WORD END END ::= {'.', '?', '!'} WORD ::= LETTER \| LETTER WORD LETTER ::= {'a'..'z', 'A'..'Z'} SPACE ::= ' ' SPACE stands for the symbol of a space. So, how many messages did Fangy send? Input The first line contains an integer n, which is the size of one message (2 ≤ n ≤ 255). The second line contains the text. The length of the text does not exceed 104 characters. It is guaranteed that the text satisfies the above described format. Specifically, this implies that the text is not empty. Output On the first and only line print the number of text messages Fangy will need. If it is impossible to split the text, print "Impossible" without the quotes. Examples Input 25 Hello. I am a little walrus. Output 2 Input 2 How are you? Output Impossible Input 19 Hello! Do you like fish? Why? Output 3 Note Let's take a look at the third sample. The text will be split into three messages: "Hello!", "Do you like fish?" and "Why?".	n = int(input()) q, w, r = [], [], [] for x in input().split("."): q += x.split("!") for x in q: w += x.split("?") for x in w: if x: r += [x.strip() + "."] i = 0 while i < len(r): if len(r[i]) > n: print("Impossible") exit() while i + 1 < len(r) and len(r[i]) + len(r[i + 1]) + 1 <= n: r[i] += "." + r[i + 1] r.pop(i + 1) i += 1 print(i)	ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR VAR VAR LIST LIST LIST FOR VAR FUNC_CALL FUNC_CALL VAR STRING VAR FUNC_CALL VAR STRING FOR VAR VAR VAR FUNC_CALL VAR STRING FOR VAR VAR IF VAR VAR LIST BIN_OP FUNC_CALL VAR STRING ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR WHILE BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP BIN_OP FUNC_CALL VAR VAR VAR FUNC_CALL VAR VAR BIN_OP VAR NUMBER NUMBER VAR VAR VAR BIN_OP STRING VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR
Fangy the little walrus, as all the modern walruses, loves to communicate via text messaging. One day he faced the following problem: When he sends large texts, they are split into parts each containing n characters (which is the size of one text message). Thus, whole sentences and words get split! Fangy did not like it, so he faced the task of breaking the text into minimal messages on his own so that no sentence were broken into pieces when it is sent and the number of text messages to be sent would be minimal. If two consecutive sentences are in different messages, the space between them can be ignored (Fangy does not write this space). The little walrus's text looks in the following manner: TEXT ::= SENTENCE \| SENTENCE SPACE TEXT SENTENCE ::= WORD SPACE SENTENCE \| WORD END END ::= {'.', '?', '!'} WORD ::= LETTER \| LETTER WORD LETTER ::= {'a'..'z', 'A'..'Z'} SPACE ::= ' ' SPACE stands for the symbol of a space. So, how many messages did Fangy send? Input The first line contains an integer n, which is the size of one message (2 ≤ n ≤ 255). The second line contains the text. The length of the text does not exceed 104 characters. It is guaranteed that the text satisfies the above described format. Specifically, this implies that the text is not empty. Output On the first and only line print the number of text messages Fangy will need. If it is impossible to split the text, print "Impossible" without the quotes. Examples Input 25 Hello. I am a little walrus. Output 2 Input 2 How are you? Output Impossible Input 19 Hello! Do you like fish? Why? Output 3 Note Let's take a look at the third sample. The text will be split into three messages: "Hello!", "Do you like fish?" and "Why?".	n = int(input()) s = input() S = [] e = "" for i in range(len(s)): if s[i] in ".?!": e += s[i] S.append(e) e = "" else: e += s[i] acc = 0 ans = 0 for item in S: x = len(item) if acc == 0: i = 0 while item[i] == " ": x -= 1 i += 1 if acc + len(item) <= n: acc += len(item) elif acc == 0: ans = -1 break else: ans += 1 acc = 0 i = 0 while item[i] == " ": x -= 1 i += 1 if x > n: ans = -1 break acc += x if ans != -1 and acc != 0: ans += 1 if ans != -1: print(ans) else: print("Impossible")	ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR STRING FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR STRING VAR VAR VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR STRING VAR NUMBER VAR NUMBER IF BIN_OP VAR FUNC_CALL VAR VAR VAR VAR FUNC_CALL VAR VAR IF VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR VAR STRING VAR NUMBER VAR NUMBER IF VAR VAR ASSIGN VAR NUMBER VAR VAR IF VAR NUMBER VAR NUMBER VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR STRING
Fangy the little walrus, as all the modern walruses, loves to communicate via text messaging. One day he faced the following problem: When he sends large texts, they are split into parts each containing n characters (which is the size of one text message). Thus, whole sentences and words get split! Fangy did not like it, so he faced the task of breaking the text into minimal messages on his own so that no sentence were broken into pieces when it is sent and the number of text messages to be sent would be minimal. If two consecutive sentences are in different messages, the space between them can be ignored (Fangy does not write this space). The little walrus's text looks in the following manner: TEXT ::= SENTENCE \| SENTENCE SPACE TEXT SENTENCE ::= WORD SPACE SENTENCE \| WORD END END ::= {'.', '?', '!'} WORD ::= LETTER \| LETTER WORD LETTER ::= {'a'..'z', 'A'..'Z'} SPACE ::= ' ' SPACE stands for the symbol of a space. So, how many messages did Fangy send? Input The first line contains an integer n, which is the size of one message (2 ≤ n ≤ 255). The second line contains the text. The length of the text does not exceed 104 characters. It is guaranteed that the text satisfies the above described format. Specifically, this implies that the text is not empty. Output On the first and only line print the number of text messages Fangy will need. If it is impossible to split the text, print "Impossible" without the quotes. Examples Input 25 Hello. I am a little walrus. Output 2 Input 2 How are you? Output Impossible Input 19 Hello! Do you like fish? Why? Output 3 Note Let's take a look at the third sample. The text will be split into three messages: "Hello!", "Do you like fish?" and "Why?".	n = int(input()) + 1 p, s, j = 0, 1, -2 for i, c in enumerate(input()): if c in ".?!": d = i - j if d > n: s = 0 break if p + d > n: s += 1 p = 0 p += d j = i print(s if s else "Impossible")	ASSIGN VAR BIN_OP FUNC_CALL VAR FUNC_CALL VAR NUMBER ASSIGN VAR VAR VAR NUMBER NUMBER NUMBER FOR VAR VAR FUNC_CALL VAR FUNC_CALL VAR IF VAR STRING ASSIGN VAR BIN_OP VAR VAR IF VAR VAR ASSIGN VAR NUMBER IF BIN_OP VAR VAR VAR VAR NUMBER ASSIGN VAR NUMBER VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR VAR STRING
Fangy the little walrus, as all the modern walruses, loves to communicate via text messaging. One day he faced the following problem: When he sends large texts, they are split into parts each containing n characters (which is the size of one text message). Thus, whole sentences and words get split! Fangy did not like it, so he faced the task of breaking the text into minimal messages on his own so that no sentence were broken into pieces when it is sent and the number of text messages to be sent would be minimal. If two consecutive sentences are in different messages, the space between them can be ignored (Fangy does not write this space). The little walrus's text looks in the following manner: TEXT ::= SENTENCE \| SENTENCE SPACE TEXT SENTENCE ::= WORD SPACE SENTENCE \| WORD END END ::= {'.', '?', '!'} WORD ::= LETTER \| LETTER WORD LETTER ::= {'a'..'z', 'A'..'Z'} SPACE ::= ' ' SPACE stands for the symbol of a space. So, how many messages did Fangy send? Input The first line contains an integer n, which is the size of one message (2 ≤ n ≤ 255). The second line contains the text. The length of the text does not exceed 104 characters. It is guaranteed that the text satisfies the above described format. Specifically, this implies that the text is not empty. Output On the first and only line print the number of text messages Fangy will need. If it is impossible to split the text, print "Impossible" without the quotes. Examples Input 25 Hello. I am a little walrus. Output 2 Input 2 How are you? Output Impossible Input 19 Hello! Do you like fish? Why? Output 3 Note Let's take a look at the third sample. The text will be split into three messages: "Hello!", "Do you like fish?" and "Why?".	n = int(input()) a = input() b = [] c = "" a1 = len(a) i = 0 while i < len(a): c += a[i] if a[i] in [".", "?", "!"]: b.append(len(c)) c = "" i += 1 i += 1 d = 0 e = 0 f = 0 for i in b: if e + i + f > n: if e == 0 or i > n: print("Impossible") break e = 0 f = 1 d += 1 else: f += 1 e += i else: print(d + 1)	ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST ASSIGN VAR STRING ASSIGN VAR FUNC_CALL VAR VAR ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR VAR VAR VAR IF VAR VAR LIST STRING STRING STRING EXPR FUNC_CALL VAR FUNC_CALL VAR VAR ASSIGN VAR STRING VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF BIN_OP BIN_OP VAR VAR VAR VAR IF VAR NUMBER VAR VAR EXPR FUNC_CALL VAR STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER VAR NUMBER VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR NUMBER
Fangy the little walrus, as all the modern walruses, loves to communicate via text messaging. One day he faced the following problem: When he sends large texts, they are split into parts each containing n characters (which is the size of one text message). Thus, whole sentences and words get split! Fangy did not like it, so he faced the task of breaking the text into minimal messages on his own so that no sentence were broken into pieces when it is sent and the number of text messages to be sent would be minimal. If two consecutive sentences are in different messages, the space between them can be ignored (Fangy does not write this space). The little walrus's text looks in the following manner: TEXT ::= SENTENCE \| SENTENCE SPACE TEXT SENTENCE ::= WORD SPACE SENTENCE \| WORD END END ::= {'.', '?', '!'} WORD ::= LETTER \| LETTER WORD LETTER ::= {'a'..'z', 'A'..'Z'} SPACE ::= ' ' SPACE stands for the symbol of a space. So, how many messages did Fangy send? Input The first line contains an integer n, which is the size of one message (2 ≤ n ≤ 255). The second line contains the text. The length of the text does not exceed 104 characters. It is guaranteed that the text satisfies the above described format. Specifically, this implies that the text is not empty. Output On the first and only line print the number of text messages Fangy will need. If it is impossible to split the text, print "Impossible" without the quotes. Examples Input 25 Hello. I am a little walrus. Output 2 Input 2 How are you? Output Impossible Input 19 Hello! Do you like fish? Why? Output 3 Note Let's take a look at the third sample. The text will be split into three messages: "Hello!", "Do you like fish?" and "Why?".	n = int(input()) text = input() text = text.replace("?", ".").replace("!", ".").split(".") ans = True qnt = 0 acu = n + 1 for i in range(len(text)): if len(text[i]) == 0: continue if len(text[i]) > n: ans = False break acu += len(text[i]) + 1 if acu > n: acu = len(text[i]) + 1 if text[i][0] == " ": acu -= 1 qnt += 1 if ans == False: print("Impossible") else: print(qnt)	ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR STRING STRING STRING STRING STRING ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR NUMBER IF FUNC_CALL VAR VAR VAR VAR ASSIGN VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER IF VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER IF VAR VAR NUMBER STRING VAR NUMBER VAR NUMBER IF VAR NUMBER EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR VAR
Fangy the little walrus, as all the modern walruses, loves to communicate via text messaging. One day he faced the following problem: When he sends large texts, they are split into parts each containing n characters (which is the size of one text message). Thus, whole sentences and words get split! Fangy did not like it, so he faced the task of breaking the text into minimal messages on his own so that no sentence were broken into pieces when it is sent and the number of text messages to be sent would be minimal. If two consecutive sentences are in different messages, the space between them can be ignored (Fangy does not write this space). The little walrus's text looks in the following manner: TEXT ::= SENTENCE \| SENTENCE SPACE TEXT SENTENCE ::= WORD SPACE SENTENCE \| WORD END END ::= {'.', '?', '!'} WORD ::= LETTER \| LETTER WORD LETTER ::= {'a'..'z', 'A'..'Z'} SPACE ::= ' ' SPACE stands for the symbol of a space. So, how many messages did Fangy send? Input The first line contains an integer n, which is the size of one message (2 ≤ n ≤ 255). The second line contains the text. The length of the text does not exceed 104 characters. It is guaranteed that the text satisfies the above described format. Specifically, this implies that the text is not empty. Output On the first and only line print the number of text messages Fangy will need. If it is impossible to split the text, print "Impossible" without the quotes. Examples Input 25 Hello. I am a little walrus. Output 2 Input 2 How are you? Output Impossible Input 19 Hello! Do you like fish? Why? Output 3 Note Let's take a look at the third sample. The text will be split into three messages: "Hello!", "Do you like fish?" and "Why?".	n = int(input()) raw = input().split(" ") sentences = [] punctuation = {"?", ".", "!"} curr = "" for idx in raw: curr += idx if idx[len(idx) - 1] in punctuation: sentences.append(curr) curr = "" else: curr += " " bad = False for idx in sentences: if len(idx) > n: bad = True if bad: print("Impossible") else: currLen = len(sentences[0]) res = 1 for idx in range(1, len(sentences)): if len(sentences[idx]) + currLen + 1 > n: res += 1 currLen = len(sentences[idx]) else: currLen += len(sentences[idx]) + 1 print(res)	ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL VAR STRING ASSIGN VAR LIST ASSIGN VAR STRING STRING STRING ASSIGN VAR STRING FOR VAR VAR VAR VAR IF VAR BIN_OP FUNC_CALL VAR VAR NUMBER VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR STRING VAR STRING ASSIGN VAR NUMBER FOR VAR VAR IF FUNC_CALL VAR VAR VAR ASSIGN VAR NUMBER IF VAR EXPR FUNC_CALL VAR STRING ASSIGN VAR FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF BIN_OP BIN_OP FUNC_CALL VAR VAR VAR VAR NUMBER VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER EXPR FUNC_CALL VAR VAR
Fangy the little walrus, as all the modern walruses, loves to communicate via text messaging. One day he faced the following problem: When he sends large texts, they are split into parts each containing n characters (which is the size of one text message). Thus, whole sentences and words get split! Fangy did not like it, so he faced the task of breaking the text into minimal messages on his own so that no sentence were broken into pieces when it is sent and the number of text messages to be sent would be minimal. If two consecutive sentences are in different messages, the space between them can be ignored (Fangy does not write this space). The little walrus's text looks in the following manner: TEXT ::= SENTENCE \| SENTENCE SPACE TEXT SENTENCE ::= WORD SPACE SENTENCE \| WORD END END ::= {'.', '?', '!'} WORD ::= LETTER \| LETTER WORD LETTER ::= {'a'..'z', 'A'..'Z'} SPACE ::= ' ' SPACE stands for the symbol of a space. So, how many messages did Fangy send? Input The first line contains an integer n, which is the size of one message (2 ≤ n ≤ 255). The second line contains the text. The length of the text does not exceed 104 characters. It is guaranteed that the text satisfies the above described format. Specifically, this implies that the text is not empty. Output On the first and only line print the number of text messages Fangy will need. If it is impossible to split the text, print "Impossible" without the quotes. Examples Input 25 Hello. I am a little walrus. Output 2 Input 2 How are you? Output Impossible Input 19 Hello! Do you like fish? Why? Output 3 Note Let's take a look at the third sample. The text will be split into three messages: "Hello!", "Do you like fish?" and "Why?".	class CodeforcesTask70BSolution: def __init__(self): self.result = "" self.n = 0 self.message = "" def read_input(self): self.n = int(input()) self.message = input() def process_task(self): sentences = [] s = "" for x in range(len(self.message)): if self.message[x] in [".", "?", "!"]: s += self.message[x] sentences.append(s) s = "" elif s or self.message[x] != " ": s += self.message[x] sent_len = [len(s) for s in sentences] if max(sent_len) > self.n: self.result = "Impossible" else: cnt = 1 cl = sent_len[0] for l in sent_len[1:]: if cl + l + 1 <= self.n: cl += l + 1 else: cnt += 1 cl = l self.result = str(cnt) def get_result(self): return self.result Solution = CodeforcesTask70BSolution() Solution.read_input() Solution.process_task() print(Solution.get_result())	CLASS_DEF FUNC_DEF ASSIGN VAR STRING ASSIGN VAR NUMBER ASSIGN VAR STRING FUNC_DEF ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_DEF ASSIGN VAR LIST ASSIGN VAR STRING FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR LIST STRING STRING STRING VAR VAR VAR EXPR FUNC_CALL VAR VAR ASSIGN VAR STRING IF VAR VAR VAR STRING VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR IF FUNC_CALL VAR VAR VAR ASSIGN VAR STRING ASSIGN VAR NUMBER ASSIGN VAR VAR NUMBER FOR VAR VAR NUMBER IF BIN_OP BIN_OP VAR VAR NUMBER VAR VAR BIN_OP VAR NUMBER VAR NUMBER ASSIGN VAR VAR ASSIGN VAR FUNC_CALL VAR VAR FUNC_DEF RETURN VAR ASSIGN VAR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR FUNC_CALL VAR
Fangy the little walrus, as all the modern walruses, loves to communicate via text messaging. One day he faced the following problem: When he sends large texts, they are split into parts each containing n characters (which is the size of one text message). Thus, whole sentences and words get split! Fangy did not like it, so he faced the task of breaking the text into minimal messages on his own so that no sentence were broken into pieces when it is sent and the number of text messages to be sent would be minimal. If two consecutive sentences are in different messages, the space between them can be ignored (Fangy does not write this space). The little walrus's text looks in the following manner: TEXT ::= SENTENCE \| SENTENCE SPACE TEXT SENTENCE ::= WORD SPACE SENTENCE \| WORD END END ::= {'.', '?', '!'} WORD ::= LETTER \| LETTER WORD LETTER ::= {'a'..'z', 'A'..'Z'} SPACE ::= ' ' SPACE stands for the symbol of a space. So, how many messages did Fangy send? Input The first line contains an integer n, which is the size of one message (2 ≤ n ≤ 255). The second line contains the text. The length of the text does not exceed 104 characters. It is guaranteed that the text satisfies the above described format. Specifically, this implies that the text is not empty. Output On the first and only line print the number of text messages Fangy will need. If it is impossible to split the text, print "Impossible" without the quotes. Examples Input 25 Hello. I am a little walrus. Output 2 Input 2 How are you? Output Impossible Input 19 Hello! Do you like fish? Why? Output 3 Note Let's take a look at the third sample. The text will be split into three messages: "Hello!", "Do you like fish?" and "Why?".	u = input e = 1 while e: n = int(u()) s = u().replace("?", ".").replace("!", ".") f = 1 p = len(s) - 1 i = 0 if "." != s[p]: f = 0 else: l = s[:p].split(".") q, t, i = 0, 1, 0 while i < len(l): r = len(l[i]) + 1 if len(str(l[i]).lstrip()) > n: f = 0 break if q + r <= n: q += r else: l[i] = str(l[i]).lstrip() q = len(l[i]) + 1 t += 1 i += 1 if f: print(t) else: print("Impossible") e -= 1	ASSIGN VAR VAR ASSIGN VAR NUMBER WHILE VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL FUNC_CALL FUNC_CALL VAR STRING STRING STRING STRING ASSIGN VAR NUMBER ASSIGN VAR BIN_OP FUNC_CALL VAR VAR NUMBER ASSIGN VAR NUMBER IF STRING VAR VAR ASSIGN VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR STRING ASSIGN VAR VAR VAR NUMBER NUMBER NUMBER WHILE VAR FUNC_CALL VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER IF FUNC_CALL VAR FUNC_CALL FUNC_CALL VAR VAR VAR VAR ASSIGN VAR NUMBER IF BIN_OP VAR VAR VAR VAR VAR ASSIGN VAR VAR FUNC_CALL FUNC_CALL VAR VAR VAR ASSIGN VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER IF VAR EXPR FUNC_CALL VAR VAR EXPR FUNC_CALL VAR STRING VAR NUMBER
Fangy the little walrus, as all the modern walruses, loves to communicate via text messaging. One day he faced the following problem: When he sends large texts, they are split into parts each containing n characters (which is the size of one text message). Thus, whole sentences and words get split! Fangy did not like it, so he faced the task of breaking the text into minimal messages on his own so that no sentence were broken into pieces when it is sent and the number of text messages to be sent would be minimal. If two consecutive sentences are in different messages, the space between them can be ignored (Fangy does not write this space). The little walrus's text looks in the following manner: TEXT ::= SENTENCE \| SENTENCE SPACE TEXT SENTENCE ::= WORD SPACE SENTENCE \| WORD END END ::= {'.', '?', '!'} WORD ::= LETTER \| LETTER WORD LETTER ::= {'a'..'z', 'A'..'Z'} SPACE ::= ' ' SPACE stands for the symbol of a space. So, how many messages did Fangy send? Input The first line contains an integer n, which is the size of one message (2 ≤ n ≤ 255). The second line contains the text. The length of the text does not exceed 104 characters. It is guaranteed that the text satisfies the above described format. Specifically, this implies that the text is not empty. Output On the first and only line print the number of text messages Fangy will need. If it is impossible to split the text, print "Impossible" without the quotes. Examples Input 25 Hello. I am a little walrus. Output 2 Input 2 How are you? Output Impossible Input 19 Hello! Do you like fish? Why? Output 3 Note Let's take a look at the third sample. The text will be split into three messages: "Hello!", "Do you like fish?" and "Why?".	n = int(input()) s = input() sens = [[]] for i in s: sens[-1].append(i) if i in [".", "!", "?"]: sens.append([]) for i in range(len(sens)): if sens[i]: sens[i] = "".join(sens[i]) sens[i] = sens[i].strip() if len(sens[i]) > n: print("Impossible") exit(0) sens.pop() i = 0 ans = 0 while i < len(sens): l = len(sens[i]) while i + 1 < len(sens) and l + 1 + len(sens[i + 1]) <= n: i += 1 l += len(sens[i]) + 1 i += 1 ans += 1 print(ans)	ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR ASSIGN VAR LIST LIST FOR VAR VAR EXPR FUNC_CALL VAR NUMBER VAR IF VAR LIST STRING STRING STRING EXPR FUNC_CALL VAR LIST FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR ASSIGN VAR VAR FUNC_CALL STRING VAR VAR ASSIGN VAR VAR FUNC_CALL VAR VAR IF FUNC_CALL VAR VAR VAR VAR EXPR FUNC_CALL VAR STRING EXPR FUNC_CALL VAR NUMBER EXPR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER WHILE VAR FUNC_CALL VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR WHILE BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP BIN_OP VAR NUMBER FUNC_CALL VAR VAR BIN_OP VAR NUMBER VAR VAR NUMBER VAR BIN_OP FUNC_CALL VAR VAR VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR
Polycarpus just has been out of luck lately! As soon as he found a job in the "Binary Cat" cafe, the club got burgled. All ice-cream was stolen. On the burglary night Polycarpus kept a careful record of all club visitors. Each time a visitor entered the club, Polycarpus put down character "+" in his notes. Similarly, each time a visitor left the club, Polycarpus put character "-" in his notes. We know that all cases of going in and out happened consecutively, that is, no two events happened at the same time. Polycarpus doesn't remember whether there was somebody in the club at the moment when his shift begun and at the moment when it ended. Right now the police wonders what minimum number of distinct people Polycarpus could have seen. Assume that he sees anybody coming in or out of the club. Each person could have come in or out an arbitrary number of times. -----Input----- The only line of the input contains a sequence of characters "+" and "-", the characters are written one after another without any separators. The characters are written in the order, in which the corresponding events occurred. The given sequence has length from 1 to 300 characters, inclusive. -----Output----- Print the sought minimum number of people -----Examples----- Input +-+-+ Output 1 Input --- Output 3	s = input() low = 0 high = 0 sum = 0 for i in s: if i == "+": sum += 1 else: sum -= 1 if sum < low: low = sum if sum > high: high = sum print(high - low)	ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR STRING VAR NUMBER VAR NUMBER IF VAR VAR ASSIGN VAR VAR IF VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR
Polycarpus just has been out of luck lately! As soon as he found a job in the "Binary Cat" cafe, the club got burgled. All ice-cream was stolen. On the burglary night Polycarpus kept a careful record of all club visitors. Each time a visitor entered the club, Polycarpus put down character "+" in his notes. Similarly, each time a visitor left the club, Polycarpus put character "-" in his notes. We know that all cases of going in and out happened consecutively, that is, no two events happened at the same time. Polycarpus doesn't remember whether there was somebody in the club at the moment when his shift begun and at the moment when it ended. Right now the police wonders what minimum number of distinct people Polycarpus could have seen. Assume that he sees anybody coming in or out of the club. Each person could have come in or out an arbitrary number of times. -----Input----- The only line of the input contains a sequence of characters "+" and "-", the characters are written one after another without any separators. The characters are written in the order, in which the corresponding events occurred. The given sequence has length from 1 to 300 characters, inclusive. -----Output----- Print the sought minimum number of people -----Examples----- Input +-+-+ Output 1 Input --- Output 3	suspects = input() minSus = 0 maxSus = 0 counter = 0 for x in suspects: if x == "+": counter += 1 minSus = min(minSus, counter) maxSus = max(maxSus, counter) else: counter -= 1 minSus = min(minSus, counter) maxSus = max(maxSus, counter) print(maxSus - minSus)	ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR STRING VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR BIN_OP VAR VAR
Polycarpus just has been out of luck lately! As soon as he found a job in the "Binary Cat" cafe, the club got burgled. All ice-cream was stolen. On the burglary night Polycarpus kept a careful record of all club visitors. Each time a visitor entered the club, Polycarpus put down character "+" in his notes. Similarly, each time a visitor left the club, Polycarpus put character "-" in his notes. We know that all cases of going in and out happened consecutively, that is, no two events happened at the same time. Polycarpus doesn't remember whether there was somebody in the club at the moment when his shift begun and at the moment when it ended. Right now the police wonders what minimum number of distinct people Polycarpus could have seen. Assume that he sees anybody coming in or out of the club. Each person could have come in or out an arbitrary number of times. -----Input----- The only line of the input contains a sequence of characters "+" and "-", the characters are written one after another without any separators. The characters are written in the order, in which the corresponding events occurred. The given sequence has length from 1 to 300 characters, inclusive. -----Output----- Print the sought minimum number of people -----Examples----- Input +-+-+ Output 1 Input --- Output 3	s = input() d, f, r = 0, 0, 0 for c in s: if c == "-": f += 1 if d >= 1: d -= 1 else: r += 1 elif c == "+": d += 1 if f >= 1: f -= 1 else: r += 1 print(r)	ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR VAR NUMBER NUMBER NUMBER FOR VAR VAR IF VAR STRING VAR NUMBER IF VAR NUMBER VAR NUMBER VAR NUMBER IF VAR STRING VAR NUMBER IF VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR
Polycarpus just has been out of luck lately! As soon as he found a job in the "Binary Cat" cafe, the club got burgled. All ice-cream was stolen. On the burglary night Polycarpus kept a careful record of all club visitors. Each time a visitor entered the club, Polycarpus put down character "+" in his notes. Similarly, each time a visitor left the club, Polycarpus put character "-" in his notes. We know that all cases of going in and out happened consecutively, that is, no two events happened at the same time. Polycarpus doesn't remember whether there was somebody in the club at the moment when his shift begun and at the moment when it ended. Right now the police wonders what minimum number of distinct people Polycarpus could have seen. Assume that he sees anybody coming in or out of the club. Each person could have come in or out an arbitrary number of times. -----Input----- The only line of the input contains a sequence of characters "+" and "-", the characters are written one after another without any separators. The characters are written in the order, in which the corresponding events occurred. The given sequence has length from 1 to 300 characters, inclusive. -----Output----- Print the sought minimum number of people -----Examples----- Input +-+-+ Output 1 Input --- Output 3	a = [(1 if x == "+" else -1) for x in input()] b = list(map(lambda i: sum(a[0:i]), range(len(a) + 1))) print(max(b) - min(b))	ASSIGN VAR VAR STRING NUMBER NUMBER VAR FUNC_CALL VAR ASSIGN VAR FUNC_CALL VAR FUNC_CALL VAR FUNC_CALL VAR VAR NUMBER VAR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR
Polycarpus just has been out of luck lately! As soon as he found a job in the "Binary Cat" cafe, the club got burgled. All ice-cream was stolen. On the burglary night Polycarpus kept a careful record of all club visitors. Each time a visitor entered the club, Polycarpus put down character "+" in his notes. Similarly, each time a visitor left the club, Polycarpus put character "-" in his notes. We know that all cases of going in and out happened consecutively, that is, no two events happened at the same time. Polycarpus doesn't remember whether there was somebody in the club at the moment when his shift begun and at the moment when it ended. Right now the police wonders what minimum number of distinct people Polycarpus could have seen. Assume that he sees anybody coming in or out of the club. Each person could have come in or out an arbitrary number of times. -----Input----- The only line of the input contains a sequence of characters "+" and "-", the characters are written one after another without any separators. The characters are written in the order, in which the corresponding events occurred. The given sequence has length from 1 to 300 characters, inclusive. -----Output----- Print the sought minimum number of people -----Examples----- Input +-+-+ Output 1 Input --- Output 3	inp = input() inclb = 0 outclb = 0 seen = 0 i = 0 for i in range(0, len(inp)): if inp[i] == "-": outclb += 1 if inclb > 0: inclb -= 1 else: seen += 1 else: inclb += 1 if outclb > 0: outclb -= 1 else: seen += 1 print(seen)	ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR NUMBER VAR NUMBER VAR NUMBER VAR NUMBER IF VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR
Polycarpus just has been out of luck lately! As soon as he found a job in the "Binary Cat" cafe, the club got burgled. All ice-cream was stolen. On the burglary night Polycarpus kept a careful record of all club visitors. Each time a visitor entered the club, Polycarpus put down character "+" in his notes. Similarly, each time a visitor left the club, Polycarpus put character "-" in his notes. We know that all cases of going in and out happened consecutively, that is, no two events happened at the same time. Polycarpus doesn't remember whether there was somebody in the club at the moment when his shift begun and at the moment when it ended. Right now the police wonders what minimum number of distinct people Polycarpus could have seen. Assume that he sees anybody coming in or out of the club. Each person could have come in or out an arbitrary number of times. -----Input----- The only line of the input contains a sequence of characters "+" and "-", the characters are written one after another without any separators. The characters are written in the order, in which the corresponding events occurred. The given sequence has length from 1 to 300 characters, inclusive. -----Output----- Print the sought minimum number of people -----Examples----- Input +-+-+ Output 1 Input --- Output 3	input_var = input() counter1 = 0 counter2 = 0 for temp in input_var: if temp == "+": counter1 += 1 if counter2 > 0: counter2 -= 1 else: counter2 += 1 if counter1 > 0: counter1 -= 1 print(counter1 + counter2)	ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR STRING VAR NUMBER IF VAR NUMBER VAR NUMBER VAR NUMBER IF VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR
Polycarpus just has been out of luck lately! As soon as he found a job in the "Binary Cat" cafe, the club got burgled. All ice-cream was stolen. On the burglary night Polycarpus kept a careful record of all club visitors. Each time a visitor entered the club, Polycarpus put down character "+" in his notes. Similarly, each time a visitor left the club, Polycarpus put character "-" in his notes. We know that all cases of going in and out happened consecutively, that is, no two events happened at the same time. Polycarpus doesn't remember whether there was somebody in the club at the moment when his shift begun and at the moment when it ended. Right now the police wonders what minimum number of distinct people Polycarpus could have seen. Assume that he sees anybody coming in or out of the club. Each person could have come in or out an arbitrary number of times. -----Input----- The only line of the input contains a sequence of characters "+" and "-", the characters are written one after another without any separators. The characters are written in the order, in which the corresponding events occurred. The given sequence has length from 1 to 300 characters, inclusive. -----Output----- Print the sought minimum number of people -----Examples----- Input +-+-+ Output 1 Input --- Output 3	s = input() a, b = 0, 0 for i in range(len(s)): if s[i] == "-": a = a - 1 if a < b: b = a else: a = a + 1 a = -b ans = a for i in range(len(s)): if s[i] == "-": a = a - 1 else: a = a + 1 if a > ans: ans = a print(ans)	ASSIGN VAR FUNC_CALL VAR ASSIGN VAR VAR NUMBER NUMBER FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR VAR ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR VAR ASSIGN VAR VAR FOR VAR FUNC_CALL VAR FUNC_CALL VAR VAR IF VAR VAR STRING ASSIGN VAR BIN_OP VAR NUMBER ASSIGN VAR BIN_OP VAR NUMBER IF VAR VAR ASSIGN VAR VAR EXPR FUNC_CALL VAR VAR
Polycarpus just has been out of luck lately! As soon as he found a job in the "Binary Cat" cafe, the club got burgled. All ice-cream was stolen. On the burglary night Polycarpus kept a careful record of all club visitors. Each time a visitor entered the club, Polycarpus put down character "+" in his notes. Similarly, each time a visitor left the club, Polycarpus put character "-" in his notes. We know that all cases of going in and out happened consecutively, that is, no two events happened at the same time. Polycarpus doesn't remember whether there was somebody in the club at the moment when his shift begun and at the moment when it ended. Right now the police wonders what minimum number of distinct people Polycarpus could have seen. Assume that he sees anybody coming in or out of the club. Each person could have come in or out an arbitrary number of times. -----Input----- The only line of the input contains a sequence of characters "+" and "-", the characters are written one after another without any separators. The characters are written in the order, in which the corresponding events occurred. The given sequence has length from 1 to 300 characters, inclusive. -----Output----- Print the sought minimum number of people -----Examples----- Input +-+-+ Output 1 Input --- Output 3	s = input() mx = 0 mn = 0 cur = 0 for i in s: if i == "+": cur += 1 else: cur -= 1 mx = max(mx, cur) mn = min(mn, cur) print(abs(mx - mn))	ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR STRING VAR NUMBER VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR ASSIGN VAR FUNC_CALL VAR VAR VAR EXPR FUNC_CALL VAR FUNC_CALL VAR BIN_OP VAR VAR
Polycarpus just has been out of luck lately! As soon as he found a job in the "Binary Cat" cafe, the club got burgled. All ice-cream was stolen. On the burglary night Polycarpus kept a careful record of all club visitors. Each time a visitor entered the club, Polycarpus put down character "+" in his notes. Similarly, each time a visitor left the club, Polycarpus put character "-" in his notes. We know that all cases of going in and out happened consecutively, that is, no two events happened at the same time. Polycarpus doesn't remember whether there was somebody in the club at the moment when his shift begun and at the moment when it ended. Right now the police wonders what minimum number of distinct people Polycarpus could have seen. Assume that he sees anybody coming in or out of the club. Each person could have come in or out an arbitrary number of times. -----Input----- The only line of the input contains a sequence of characters "+" and "-", the characters are written one after another without any separators. The characters are written in the order, in which the corresponding events occurred. The given sequence has length from 1 to 300 characters, inclusive. -----Output----- Print the sought minimum number of people -----Examples----- Input +-+-+ Output 1 Input --- Output 3	t = input() inPerson = 0 outPerson = 0 for i in range(0, len(t)): if t[i] == "+": inPerson += 1 if outPerson: outPerson -= 1 elif t[i] == "-": outPerson += 1 if inPerson: inPerson -= 1 print(inPerson + outPerson)	ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR IF VAR VAR STRING VAR NUMBER IF VAR VAR NUMBER IF VAR VAR STRING VAR NUMBER IF VAR VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR
Polycarpus just has been out of luck lately! As soon as he found a job in the "Binary Cat" cafe, the club got burgled. All ice-cream was stolen. On the burglary night Polycarpus kept a careful record of all club visitors. Each time a visitor entered the club, Polycarpus put down character "+" in his notes. Similarly, each time a visitor left the club, Polycarpus put character "-" in his notes. We know that all cases of going in and out happened consecutively, that is, no two events happened at the same time. Polycarpus doesn't remember whether there was somebody in the club at the moment when his shift begun and at the moment when it ended. Right now the police wonders what minimum number of distinct people Polycarpus could have seen. Assume that he sees anybody coming in or out of the club. Each person could have come in or out an arbitrary number of times. -----Input----- The only line of the input contains a sequence of characters "+" and "-", the characters are written one after another without any separators. The characters are written in the order, in which the corresponding events occurred. The given sequence has length from 1 to 300 characters, inclusive. -----Output----- Print the sought minimum number of people -----Examples----- Input +-+-+ Output 1 Input --- Output 3	n = input() inside_count = 0 outside_count = 0 for element in n: if element == "+": inside_count += 1 outside_count -= 1 if outside_count < 0 or inside_count > 100: outside_count = 0 else: inside_count -= 1 outside_count += 1 if inside_count < 0 or outside_count > 100: inside_count = 0 print(inside_count + outside_count)	ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR STRING VAR NUMBER VAR NUMBER IF VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER VAR NUMBER VAR NUMBER IF VAR NUMBER VAR NUMBER ASSIGN VAR NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR
Polycarpus just has been out of luck lately! As soon as he found a job in the "Binary Cat" cafe, the club got burgled. All ice-cream was stolen. On the burglary night Polycarpus kept a careful record of all club visitors. Each time a visitor entered the club, Polycarpus put down character "+" in his notes. Similarly, each time a visitor left the club, Polycarpus put character "-" in his notes. We know that all cases of going in and out happened consecutively, that is, no two events happened at the same time. Polycarpus doesn't remember whether there was somebody in the club at the moment when his shift begun and at the moment when it ended. Right now the police wonders what minimum number of distinct people Polycarpus could have seen. Assume that he sees anybody coming in or out of the club. Each person could have come in or out an arbitrary number of times. -----Input----- The only line of the input contains a sequence of characters "+" and "-", the characters are written one after another without any separators. The characters are written in the order, in which the corresponding events occurred. The given sequence has length from 1 to 300 characters, inclusive. -----Output----- Print the sought minimum number of people -----Examples----- Input +-+-+ Output 1 Input --- Output 3	p = [0] + [(1 if i == "+" else -1) for i in input()] for i in range(1, len(p)): p[i] += p[i - 1] print(max(p) - min(p))	ASSIGN VAR BIN_OP LIST NUMBER VAR STRING NUMBER NUMBER VAR FUNC_CALL VAR FOR VAR FUNC_CALL VAR NUMBER FUNC_CALL VAR VAR VAR VAR VAR BIN_OP VAR NUMBER EXPR FUNC_CALL VAR BIN_OP FUNC_CALL VAR VAR FUNC_CALL VAR VAR
Polycarpus just has been out of luck lately! As soon as he found a job in the "Binary Cat" cafe, the club got burgled. All ice-cream was stolen. On the burglary night Polycarpus kept a careful record of all club visitors. Each time a visitor entered the club, Polycarpus put down character "+" in his notes. Similarly, each time a visitor left the club, Polycarpus put character "-" in his notes. We know that all cases of going in and out happened consecutively, that is, no two events happened at the same time. Polycarpus doesn't remember whether there was somebody in the club at the moment when his shift begun and at the moment when it ended. Right now the police wonders what minimum number of distinct people Polycarpus could have seen. Assume that he sees anybody coming in or out of the club. Each person could have come in or out an arbitrary number of times. -----Input----- The only line of the input contains a sequence of characters "+" and "-", the characters are written one after another without any separators. The characters are written in the order, in which the corresponding events occurred. The given sequence has length from 1 to 300 characters, inclusive. -----Output----- Print the sought minimum number of people -----Examples----- Input +-+-+ Output 1 Input --- Output 3	def main(): s = input() cnt = 0 ans = 0 for x in s: if x == "+": cnt += 1 ans = max(ans, cnt) elif cnt == 0: ans += 1 else: cnt -= 1 print(ans) def __starting_point(): main() __starting_point()	FUNC_DEF ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR STRING VAR NUMBER ASSIGN VAR FUNC_CALL VAR VAR VAR IF VAR NUMBER VAR NUMBER VAR NUMBER EXPR FUNC_CALL VAR VAR FUNC_DEF EXPR FUNC_CALL VAR EXPR FUNC_CALL VAR
Polycarpus just has been out of luck lately! As soon as he found a job in the "Binary Cat" cafe, the club got burgled. All ice-cream was stolen. On the burglary night Polycarpus kept a careful record of all club visitors. Each time a visitor entered the club, Polycarpus put down character "+" in his notes. Similarly, each time a visitor left the club, Polycarpus put character "-" in his notes. We know that all cases of going in and out happened consecutively, that is, no two events happened at the same time. Polycarpus doesn't remember whether there was somebody in the club at the moment when his shift begun and at the moment when it ended. Right now the police wonders what minimum number of distinct people Polycarpus could have seen. Assume that he sees anybody coming in or out of the club. Each person could have come in or out an arbitrary number of times. -----Input----- The only line of the input contains a sequence of characters "+" and "-", the characters are written one after another without any separators. The characters are written in the order, in which the corresponding events occurred. The given sequence has length from 1 to 300 characters, inclusive. -----Output----- Print the sought minimum number of people -----Examples----- Input +-+-+ Output 1 Input --- Output 3	s = input() p = 0 m = 0 for i in s: if i == "+": p += 1 m = max(m - 1, 0) elif i == "-": m += 1 p = max(p - 1, 0) print(p + m)	ASSIGN VAR FUNC_CALL VAR ASSIGN VAR NUMBER ASSIGN VAR NUMBER FOR VAR VAR IF VAR STRING VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER IF VAR STRING VAR NUMBER ASSIGN VAR FUNC_CALL VAR BIN_OP VAR NUMBER NUMBER EXPR FUNC_CALL VAR BIN_OP VAR VAR

Subsets and Splits

No community queries yet

The top public SQL queries from the community will appear here once available.