message stringlengths 2 65.1k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 0 108k | cluster float64 14 14 | __index_level_0__ int64 0 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
As you have noticed, there are lovely girls in Arpa’s land.
People in Arpa's land are numbered from 1 to n. Everyone has exactly one crush, i-th person's crush is person with the number crushi.
<image>
Someday Arpa shouted Owf loudly from the top of the palace and a funny game started in Arpa's land. The rules are as follows.
The game consists of rounds. Assume person x wants to start a round, he calls crushx and says: "Oww...wwf" (the letter w is repeated t times) and cuts off the phone immediately. If t > 1 then crushx calls crushcrushx and says: "Oww...wwf" (the letter w is repeated t - 1 times) and cuts off the phone immediately. The round continues until some person receives an "Owf" (t = 1). This person is called the Joon-Joon of the round. There can't be two rounds at the same time.
Mehrdad has an evil plan to make the game more funny, he wants to find smallest t (t ≥ 1) such that for each person x, if x starts some round and y becomes the Joon-Joon of the round, then by starting from y, x would become the Joon-Joon of the round. Find such t for Mehrdad if it's possible.
Some strange fact in Arpa's land is that someone can be himself's crush (i.e. crushi = i).
Input
The first line of input contains integer n (1 ≤ n ≤ 100) — the number of people in Arpa's land.
The second line contains n integers, i-th of them is crushi (1 ≤ crushi ≤ n) — the number of i-th person's crush.
Output
If there is no t satisfying the condition, print -1. Otherwise print such smallest t.
Examples
Input
4
2 3 1 4
Output
3
Input
4
4 4 4 4
Output
-1
Input
4
2 1 4 3
Output
1
Note
In the first sample suppose t = 3.
If the first person starts some round:
The first person calls the second person and says "Owwwf", then the second person calls the third person and says "Owwf", then the third person calls the first person and says "Owf", so the first person becomes Joon-Joon of the round. So the condition is satisfied if x is 1.
The process is similar for the second and the third person.
If the fourth person starts some round:
The fourth person calls himself and says "Owwwf", then he calls himself again and says "Owwf", then he calls himself for another time and says "Owf", so the fourth person becomes Joon-Joon of the round. So the condition is satisfied when x is 4.
In the last example if the first person starts a round, then the second person becomes the Joon-Joon, and vice versa. | instruction | 0 | 99,141 | 14 | 198,282 |
Tags: dfs and similar, math
Correct Solution:
```
def gcd(a, b):
return a if b == 0 else gcd(b, a % b)
def lcm(a, b):
return a * b // gcd(a, b)
n = int(input())
c = [int(v)-1 for v in input().split(' ')]
t = 1
for i in range(n):
k = i
for j in range(1, n+2):
k = c[k]
if k == i:
break
if j > n:
print(-1)
exit()
t = lcm(t, j if j % 2 != 0 else j // 2)
print(t)
# Made By Mostafa_Khaled
``` | output | 1 | 99,141 | 14 | 198,283 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you have noticed, there are lovely girls in Arpa’s land.
People in Arpa's land are numbered from 1 to n. Everyone has exactly one crush, i-th person's crush is person with the number crushi.
<image>
Someday Arpa shouted Owf loudly from the top of the palace and a funny game started in Arpa's land. The rules are as follows.
The game consists of rounds. Assume person x wants to start a round, he calls crushx and says: "Oww...wwf" (the letter w is repeated t times) and cuts off the phone immediately. If t > 1 then crushx calls crushcrushx and says: "Oww...wwf" (the letter w is repeated t - 1 times) and cuts off the phone immediately. The round continues until some person receives an "Owf" (t = 1). This person is called the Joon-Joon of the round. There can't be two rounds at the same time.
Mehrdad has an evil plan to make the game more funny, he wants to find smallest t (t ≥ 1) such that for each person x, if x starts some round and y becomes the Joon-Joon of the round, then by starting from y, x would become the Joon-Joon of the round. Find such t for Mehrdad if it's possible.
Some strange fact in Arpa's land is that someone can be himself's crush (i.e. crushi = i).
Input
The first line of input contains integer n (1 ≤ n ≤ 100) — the number of people in Arpa's land.
The second line contains n integers, i-th of them is crushi (1 ≤ crushi ≤ n) — the number of i-th person's crush.
Output
If there is no t satisfying the condition, print -1. Otherwise print such smallest t.
Examples
Input
4
2 3 1 4
Output
3
Input
4
4 4 4 4
Output
-1
Input
4
2 1 4 3
Output
1
Note
In the first sample suppose t = 3.
If the first person starts some round:
The first person calls the second person and says "Owwwf", then the second person calls the third person and says "Owwf", then the third person calls the first person and says "Owf", so the first person becomes Joon-Joon of the round. So the condition is satisfied if x is 1.
The process is similar for the second and the third person.
If the fourth person starts some round:
The fourth person calls himself and says "Owwwf", then he calls himself again and says "Owwf", then he calls himself for another time and says "Owf", so the fourth person becomes Joon-Joon of the round. So the condition is satisfied when x is 4.
In the last example if the first person starts a round, then the second person becomes the Joon-Joon, and vice versa.
Submitted Solution:
```
import math
n=int(input())
a=list(map(int,input().split()))
if len(set(a))!=len(a): print(-1)
else:
ans,cycle=None,None
visited=[False]*(n+1)
for i in range(n):
if visited[i+1]: continue
count,v=0,[]
index,value=i+1,-1
while value!=i+1:
visited[index]=True
v.append(index)
value=a[index-1]
index=value
cycle=[len(v)//2,len(v)][len(v)%2]
if i==0: ans=cycle
else: ans=(ans*cycle)//math.gcd(ans,cycle)
print(ans)
``` | instruction | 0 | 99,142 | 14 | 198,284 |
Yes | output | 1 | 99,142 | 14 | 198,285 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you have noticed, there are lovely girls in Arpa’s land.
People in Arpa's land are numbered from 1 to n. Everyone has exactly one crush, i-th person's crush is person with the number crushi.
<image>
Someday Arpa shouted Owf loudly from the top of the palace and a funny game started in Arpa's land. The rules are as follows.
The game consists of rounds. Assume person x wants to start a round, he calls crushx and says: "Oww...wwf" (the letter w is repeated t times) and cuts off the phone immediately. If t > 1 then crushx calls crushcrushx and says: "Oww...wwf" (the letter w is repeated t - 1 times) and cuts off the phone immediately. The round continues until some person receives an "Owf" (t = 1). This person is called the Joon-Joon of the round. There can't be two rounds at the same time.
Mehrdad has an evil plan to make the game more funny, he wants to find smallest t (t ≥ 1) such that for each person x, if x starts some round and y becomes the Joon-Joon of the round, then by starting from y, x would become the Joon-Joon of the round. Find such t for Mehrdad if it's possible.
Some strange fact in Arpa's land is that someone can be himself's crush (i.e. crushi = i).
Input
The first line of input contains integer n (1 ≤ n ≤ 100) — the number of people in Arpa's land.
The second line contains n integers, i-th of them is crushi (1 ≤ crushi ≤ n) — the number of i-th person's crush.
Output
If there is no t satisfying the condition, print -1. Otherwise print such smallest t.
Examples
Input
4
2 3 1 4
Output
3
Input
4
4 4 4 4
Output
-1
Input
4
2 1 4 3
Output
1
Note
In the first sample suppose t = 3.
If the first person starts some round:
The first person calls the second person and says "Owwwf", then the second person calls the third person and says "Owwf", then the third person calls the first person and says "Owf", so the first person becomes Joon-Joon of the round. So the condition is satisfied if x is 1.
The process is similar for the second and the third person.
If the fourth person starts some round:
The fourth person calls himself and says "Owwwf", then he calls himself again and says "Owwf", then he calls himself for another time and says "Owf", so the fourth person becomes Joon-Joon of the round. So the condition is satisfied when x is 4.
In the last example if the first person starts a round, then the second person becomes the Joon-Joon, and vice versa.
Submitted Solution:
```
from fractions import gcd
def dfs(node):
if (vis[node] == idd[0]):
return node
cycleLen[0] += 1
vis[node] = idd[0]
return dfs(crush[node])
n = int(input())
crush = list(map(int, input().split()))
for i in range(n):
crush[i] -= 1
vis = [0] * n
ans = 10**100
idd = [0]
V = []
for i in range(n):
idd[0] += 1
cycleLen = [0]
if (dfs(i) != i):
ans = -1
V.append(cycleLen[0])
V.sort()
def lcm(x, y):
return x * y // gcd(x, y)
if (ans == -1):
print(ans)
else:
ans = 1
inc = 1
for item in V:
mem = {}
bad = False
while (True):
if (ans % item == 0):
break
if (item % 2 == 0 and ans % item == item // 2):
break
if (ans % item in mem):
bad = True
break
mem[ans % item] = 1
ans += inc
if (bad):
ans = -1
break
inc = lcm(inc, ans)
print(ans)
``` | instruction | 0 | 99,143 | 14 | 198,286 |
Yes | output | 1 | 99,143 | 14 | 198,287 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you have noticed, there are lovely girls in Arpa’s land.
People in Arpa's land are numbered from 1 to n. Everyone has exactly one crush, i-th person's crush is person with the number crushi.
<image>
Someday Arpa shouted Owf loudly from the top of the palace and a funny game started in Arpa's land. The rules are as follows.
The game consists of rounds. Assume person x wants to start a round, he calls crushx and says: "Oww...wwf" (the letter w is repeated t times) and cuts off the phone immediately. If t > 1 then crushx calls crushcrushx and says: "Oww...wwf" (the letter w is repeated t - 1 times) and cuts off the phone immediately. The round continues until some person receives an "Owf" (t = 1). This person is called the Joon-Joon of the round. There can't be two rounds at the same time.
Mehrdad has an evil plan to make the game more funny, he wants to find smallest t (t ≥ 1) such that for each person x, if x starts some round and y becomes the Joon-Joon of the round, then by starting from y, x would become the Joon-Joon of the round. Find such t for Mehrdad if it's possible.
Some strange fact in Arpa's land is that someone can be himself's crush (i.e. crushi = i).
Input
The first line of input contains integer n (1 ≤ n ≤ 100) — the number of people in Arpa's land.
The second line contains n integers, i-th of them is crushi (1 ≤ crushi ≤ n) — the number of i-th person's crush.
Output
If there is no t satisfying the condition, print -1. Otherwise print such smallest t.
Examples
Input
4
2 3 1 4
Output
3
Input
4
4 4 4 4
Output
-1
Input
4
2 1 4 3
Output
1
Note
In the first sample suppose t = 3.
If the first person starts some round:
The first person calls the second person and says "Owwwf", then the second person calls the third person and says "Owwf", then the third person calls the first person and says "Owf", so the first person becomes Joon-Joon of the round. So the condition is satisfied if x is 1.
The process is similar for the second and the third person.
If the fourth person starts some round:
The fourth person calls himself and says "Owwwf", then he calls himself again and says "Owwf", then he calls himself for another time and says "Owf", so the fourth person becomes Joon-Joon of the round. So the condition is satisfied when x is 4.
In the last example if the first person starts a round, then the second person becomes the Joon-Joon, and vice versa.
Submitted Solution:
```
from sys import stdin, stdout, setrecursionlimit
input = stdin.readline
# setrecursionlimit(int(1e6))
inf = float('inf')
from collections import defaultdict as dd
from collections import Counter, deque
from heapq import *
import math
from math import floor, ceil, sqrt
def geti(): return map(int, input().strip().split())
def getl(): return list(map(int, input().strip().split()))
def getis(): return map(str, input().strip().split())
def getls(): return list(map(str, input().strip().split()))
def gets(): return input().strip()
def geta(): return int(input())
def print_s(s): stdout.write(s+'\n')
def solve():
n = geta()
a = getl()
edges = dd(list)
for i in range(n):
edges[i].append(a[i]-1)
dist = [[inf]*n for _ in range(n)]
def dfs(root, par, temp = 0):
vis[root] = True
for node in edges[root]:
dist[par][node] = temp + 1
if node not in vis:
dfs(node, par, temp+1)
ans = []
vis = dd(int)
ok = dd(int)
def dfs(root, par, dist = 0):
if root == par and dist:
ok[par] = True
if dist&1:
ans.append(dist)
else:
ans.append(dist//2)
return
if root in vis:
return
vis[root] = True
for node in edges[root]:
dfs(node, par, dist+1)
need = []
for i in range(n):
if i not in vis:
need.append(i)
dfs(i, i)
# print(ans)
for i in need:
if i not in ok:
print(-1)
break
else:
ans = Counter(ans)
ans = sorted(ans)
# print(ans)
vis = [False]*(len(ans)+1)
ok = []
for i in range(len(ans)):
if ans[i] == 1 or vis[i]:
continue
temp = []
for j in range(i, len(ans)):
if ans[j] % ans[i] == 0:
temp.append(ans[j])
vis[j] = True
now = temp[0]
g = temp[0]
# print(temp)
for k in range(1, len(temp)):
now *= temp[k]
g = math.gcd(g, temp[k])
now = now//g
g = now
if g!=now:
ok.append(now//g)
else:
ok.append(now)
# print(ok)
now = 1
if ok:
now = ok[0]
g = ok[0]
for k in range(1, len(ok)):
now *= ok[k]
g = math.gcd(g, ok[k])
now = now//g
g = now
print(now)
if __name__=='__main__':
solve()
``` | instruction | 0 | 99,144 | 14 | 198,288 |
Yes | output | 1 | 99,144 | 14 | 198,289 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you have noticed, there are lovely girls in Arpa’s land.
People in Arpa's land are numbered from 1 to n. Everyone has exactly one crush, i-th person's crush is person with the number crushi.
<image>
Someday Arpa shouted Owf loudly from the top of the palace and a funny game started in Arpa's land. The rules are as follows.
The game consists of rounds. Assume person x wants to start a round, he calls crushx and says: "Oww...wwf" (the letter w is repeated t times) and cuts off the phone immediately. If t > 1 then crushx calls crushcrushx and says: "Oww...wwf" (the letter w is repeated t - 1 times) and cuts off the phone immediately. The round continues until some person receives an "Owf" (t = 1). This person is called the Joon-Joon of the round. There can't be two rounds at the same time.
Mehrdad has an evil plan to make the game more funny, he wants to find smallest t (t ≥ 1) such that for each person x, if x starts some round and y becomes the Joon-Joon of the round, then by starting from y, x would become the Joon-Joon of the round. Find such t for Mehrdad if it's possible.
Some strange fact in Arpa's land is that someone can be himself's crush (i.e. crushi = i).
Input
The first line of input contains integer n (1 ≤ n ≤ 100) — the number of people in Arpa's land.
The second line contains n integers, i-th of them is crushi (1 ≤ crushi ≤ n) — the number of i-th person's crush.
Output
If there is no t satisfying the condition, print -1. Otherwise print such smallest t.
Examples
Input
4
2 3 1 4
Output
3
Input
4
4 4 4 4
Output
-1
Input
4
2 1 4 3
Output
1
Note
In the first sample suppose t = 3.
If the first person starts some round:
The first person calls the second person and says "Owwwf", then the second person calls the third person and says "Owwf", then the third person calls the first person and says "Owf", so the first person becomes Joon-Joon of the round. So the condition is satisfied if x is 1.
The process is similar for the second and the third person.
If the fourth person starts some round:
The fourth person calls himself and says "Owwwf", then he calls himself again and says "Owwf", then he calls himself for another time and says "Owf", so the fourth person becomes Joon-Joon of the round. So the condition is satisfied when x is 4.
In the last example if the first person starts a round, then the second person becomes the Joon-Joon, and vice versa.
Submitted Solution:
```
#from collections import deque
from functools import reduce
n = int(input())
crush = [int(i) - 1 for i in input().split()]
def parity_treat(n):
if n%2 == 0:
return n//2
else:
return n
def gcd(a,b):
while b:
a, b = b, a%b
return a
def lcm(a,b):
return a * b // gcd(a,b)
def lcmm(*args):
return reduce(lcm, args)
if len(set(crush)) < n:
print(-1)
else:
component_size = []
visited = set()
for i in range(n):
if i not in visited:
tmp = 1
start = i
visited.add(start)
j = crush[start]
while j != start:
visited.add(j)
j = crush[j]
tmp+=1
component_size.append(tmp)
component_size = [parity_treat(i) for i in component_size]
print(lcmm(*component_size))
``` | instruction | 0 | 99,145 | 14 | 198,290 |
Yes | output | 1 | 99,145 | 14 | 198,291 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you have noticed, there are lovely girls in Arpa’s land.
People in Arpa's land are numbered from 1 to n. Everyone has exactly one crush, i-th person's crush is person with the number crushi.
<image>
Someday Arpa shouted Owf loudly from the top of the palace and a funny game started in Arpa's land. The rules are as follows.
The game consists of rounds. Assume person x wants to start a round, he calls crushx and says: "Oww...wwf" (the letter w is repeated t times) and cuts off the phone immediately. If t > 1 then crushx calls crushcrushx and says: "Oww...wwf" (the letter w is repeated t - 1 times) and cuts off the phone immediately. The round continues until some person receives an "Owf" (t = 1). This person is called the Joon-Joon of the round. There can't be two rounds at the same time.
Mehrdad has an evil plan to make the game more funny, he wants to find smallest t (t ≥ 1) such that for each person x, if x starts some round and y becomes the Joon-Joon of the round, then by starting from y, x would become the Joon-Joon of the round. Find such t for Mehrdad if it's possible.
Some strange fact in Arpa's land is that someone can be himself's crush (i.e. crushi = i).
Input
The first line of input contains integer n (1 ≤ n ≤ 100) — the number of people in Arpa's land.
The second line contains n integers, i-th of them is crushi (1 ≤ crushi ≤ n) — the number of i-th person's crush.
Output
If there is no t satisfying the condition, print -1. Otherwise print such smallest t.
Examples
Input
4
2 3 1 4
Output
3
Input
4
4 4 4 4
Output
-1
Input
4
2 1 4 3
Output
1
Note
In the first sample suppose t = 3.
If the first person starts some round:
The first person calls the second person and says "Owwwf", then the second person calls the third person and says "Owwf", then the third person calls the first person and says "Owf", so the first person becomes Joon-Joon of the round. So the condition is satisfied if x is 1.
The process is similar for the second and the third person.
If the fourth person starts some round:
The fourth person calls himself and says "Owwwf", then he calls himself again and says "Owwf", then he calls himself for another time and says "Owf", so the fourth person becomes Joon-Joon of the round. So the condition is satisfied when x is 4.
In the last example if the first person starts a round, then the second person becomes the Joon-Joon, and vice versa.
Submitted Solution:
```
def gcd(a,b):
while b > 0:
a, b = b, a % b
return a
def lcm(a, b):
return int(a * b / gcd(a, b))
def run(n, crush):
visited = [False] * n
cycle_size = 1
for i in range(0, n):
if visited[i]:
continue
x = i
c = 0
while (not visited[x]):
visited[x] = True
x = crush[x] - 1
c += 1
if x != i:
return -1
cycle_size = lcm(cycle_size, c)
if cycle_size % 2 == 0:
cycle_size /= 2
return cycle_size
n = int(input())
crush = [int(x) for x in input().split()]
print(run(n,crush))
``` | instruction | 0 | 99,146 | 14 | 198,292 |
No | output | 1 | 99,146 | 14 | 198,293 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you have noticed, there are lovely girls in Arpa’s land.
People in Arpa's land are numbered from 1 to n. Everyone has exactly one crush, i-th person's crush is person with the number crushi.
<image>
Someday Arpa shouted Owf loudly from the top of the palace and a funny game started in Arpa's land. The rules are as follows.
The game consists of rounds. Assume person x wants to start a round, he calls crushx and says: "Oww...wwf" (the letter w is repeated t times) and cuts off the phone immediately. If t > 1 then crushx calls crushcrushx and says: "Oww...wwf" (the letter w is repeated t - 1 times) and cuts off the phone immediately. The round continues until some person receives an "Owf" (t = 1). This person is called the Joon-Joon of the round. There can't be two rounds at the same time.
Mehrdad has an evil plan to make the game more funny, he wants to find smallest t (t ≥ 1) such that for each person x, if x starts some round and y becomes the Joon-Joon of the round, then by starting from y, x would become the Joon-Joon of the round. Find such t for Mehrdad if it's possible.
Some strange fact in Arpa's land is that someone can be himself's crush (i.e. crushi = i).
Input
The first line of input contains integer n (1 ≤ n ≤ 100) — the number of people in Arpa's land.
The second line contains n integers, i-th of them is crushi (1 ≤ crushi ≤ n) — the number of i-th person's crush.
Output
If there is no t satisfying the condition, print -1. Otherwise print such smallest t.
Examples
Input
4
2 3 1 4
Output
3
Input
4
4 4 4 4
Output
-1
Input
4
2 1 4 3
Output
1
Note
In the first sample suppose t = 3.
If the first person starts some round:
The first person calls the second person and says "Owwwf", then the second person calls the third person and says "Owwf", then the third person calls the first person and says "Owf", so the first person becomes Joon-Joon of the round. So the condition is satisfied if x is 1.
The process is similar for the second and the third person.
If the fourth person starts some round:
The fourth person calls himself and says "Owwwf", then he calls himself again and says "Owwf", then he calls himself for another time and says "Owf", so the fourth person becomes Joon-Joon of the round. So the condition is satisfied when x is 4.
In the last example if the first person starts a round, then the second person becomes the Joon-Joon, and vice versa.
Submitted Solution:
```
###### ### ####### ####### ## # ##### ### #####
# # # # # # # # # # # # # ###
# # # # # # # # # # # # # ###
###### ######### # # # # # # ######### #
###### ######### # # # # # # ######### #
# # # # # # # # # # #### # # #
# # # # # # # ## # # # # #
###### # # ####### ####### # # ##### # # # #
from __future__ import print_function # for PyPy2
# from itertools import permutations
# from functools import cmp_to_key # for adding custom comparator
# from fractions import Fraction
from collections import *
from sys import stdin
# from bisect import *
from heapq import *
from math import *
g = lambda : stdin.readline().strip()
gl = lambda : g().split()
gil = lambda : [int(var) for var in gl()]
gfl = lambda : [float(var) for var in gl()]
gcl = lambda : list(g())
gbs = lambda : [int(var) for var in g()]
rr = lambda x : reversed(range(x))
mod = int(1e9)+7
inf = float("inf")
n, = gil()
nxt = [0] + gil()
vis = [0]*(n+1)
c = set()
def cycle(p):
tt = 0
start = p
while vis[p] == 0:
tt += 1
vis[p] = 1
p = nxt[p]
if p != start:
print(-1)
exit()
return tt
for p in range(1, n+1):
if vis[p] : continue
tt = cycle(p)
if tt&1 :
c.add(tt)
else:
c.add(tt//2)
if 1 in c:c.remove(1)
c = list(c)
if len(c) <= 1:
print(c[0] if c else 1)
exit()
p = 1
gc = c[0] if c else 1
for ci in c:
gc = gcd(gc, ci)
p *= ci
# print(p, gc)
print(p//gc)
``` | instruction | 0 | 99,147 | 14 | 198,294 |
No | output | 1 | 99,147 | 14 | 198,295 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you have noticed, there are lovely girls in Arpa’s land.
People in Arpa's land are numbered from 1 to n. Everyone has exactly one crush, i-th person's crush is person with the number crushi.
<image>
Someday Arpa shouted Owf loudly from the top of the palace and a funny game started in Arpa's land. The rules are as follows.
The game consists of rounds. Assume person x wants to start a round, he calls crushx and says: "Oww...wwf" (the letter w is repeated t times) and cuts off the phone immediately. If t > 1 then crushx calls crushcrushx and says: "Oww...wwf" (the letter w is repeated t - 1 times) and cuts off the phone immediately. The round continues until some person receives an "Owf" (t = 1). This person is called the Joon-Joon of the round. There can't be two rounds at the same time.
Mehrdad has an evil plan to make the game more funny, he wants to find smallest t (t ≥ 1) such that for each person x, if x starts some round and y becomes the Joon-Joon of the round, then by starting from y, x would become the Joon-Joon of the round. Find such t for Mehrdad if it's possible.
Some strange fact in Arpa's land is that someone can be himself's crush (i.e. crushi = i).
Input
The first line of input contains integer n (1 ≤ n ≤ 100) — the number of people in Arpa's land.
The second line contains n integers, i-th of them is crushi (1 ≤ crushi ≤ n) — the number of i-th person's crush.
Output
If there is no t satisfying the condition, print -1. Otherwise print such smallest t.
Examples
Input
4
2 3 1 4
Output
3
Input
4
4 4 4 4
Output
-1
Input
4
2 1 4 3
Output
1
Note
In the first sample suppose t = 3.
If the first person starts some round:
The first person calls the second person and says "Owwwf", then the second person calls the third person and says "Owwf", then the third person calls the first person and says "Owf", so the first person becomes Joon-Joon of the round. So the condition is satisfied if x is 1.
The process is similar for the second and the third person.
If the fourth person starts some round:
The fourth person calls himself and says "Owwwf", then he calls himself again and says "Owwf", then he calls himself for another time and says "Owf", so the fourth person becomes Joon-Joon of the round. So the condition is satisfied when x is 4.
In the last example if the first person starts a round, then the second person becomes the Joon-Joon, and vice versa.
Submitted Solution:
```
input()
crush = [0] + [int(x) for x in input().split()]
visited = set()
circle_sizes = []
def solve():
for i in range(len(crush)):
if i not in visited:
start, cur, count = i, i, 0
while cur not in visited:
visited.add(cur)
count += 1
cur = crush[cur]
if cur != start:
return -1
circle_sizes.append(count if count % 2 else count // 2)
ans = max(circle_sizes)
for size in circle_sizes:
if ans % size != 0:
return -1
return ans
print(solve())
``` | instruction | 0 | 99,148 | 14 | 198,296 |
No | output | 1 | 99,148 | 14 | 198,297 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you have noticed, there are lovely girls in Arpa’s land.
People in Arpa's land are numbered from 1 to n. Everyone has exactly one crush, i-th person's crush is person with the number crushi.
<image>
Someday Arpa shouted Owf loudly from the top of the palace and a funny game started in Arpa's land. The rules are as follows.
The game consists of rounds. Assume person x wants to start a round, he calls crushx and says: "Oww...wwf" (the letter w is repeated t times) and cuts off the phone immediately. If t > 1 then crushx calls crushcrushx and says: "Oww...wwf" (the letter w is repeated t - 1 times) and cuts off the phone immediately. The round continues until some person receives an "Owf" (t = 1). This person is called the Joon-Joon of the round. There can't be two rounds at the same time.
Mehrdad has an evil plan to make the game more funny, he wants to find smallest t (t ≥ 1) such that for each person x, if x starts some round and y becomes the Joon-Joon of the round, then by starting from y, x would become the Joon-Joon of the round. Find such t for Mehrdad if it's possible.
Some strange fact in Arpa's land is that someone can be himself's crush (i.e. crushi = i).
Input
The first line of input contains integer n (1 ≤ n ≤ 100) — the number of people in Arpa's land.
The second line contains n integers, i-th of them is crushi (1 ≤ crushi ≤ n) — the number of i-th person's crush.
Output
If there is no t satisfying the condition, print -1. Otherwise print such smallest t.
Examples
Input
4
2 3 1 4
Output
3
Input
4
4 4 4 4
Output
-1
Input
4
2 1 4 3
Output
1
Note
In the first sample suppose t = 3.
If the first person starts some round:
The first person calls the second person and says "Owwwf", then the second person calls the third person and says "Owwf", then the third person calls the first person and says "Owf", so the first person becomes Joon-Joon of the round. So the condition is satisfied if x is 1.
The process is similar for the second and the third person.
If the fourth person starts some round:
The fourth person calls himself and says "Owwwf", then he calls himself again and says "Owwf", then he calls himself for another time and says "Owf", so the fourth person becomes Joon-Joon of the round. So the condition is satisfied when x is 4.
In the last example if the first person starts a round, then the second person becomes the Joon-Joon, and vice versa.
Submitted Solution:
```
def read_int():
return int(input())
def read_ints():
return map(int, input().split(' '))
def gcd(x, y):
return x if y == 0 else gcd(y, x % y)
n = read_int()
a = list(read_ints())
vis = [False for i in range(n)]
loop = 0
ok = True
for i in range(n):
if vis[i]:
continue
current = i
length = 0
while not vis[current]:
vis[current] = True
current = a[current] - 1
length += 1
if current != i:
ok = False
print(-1)
break
if loop == 0:
loop = length
else:
loop = loop * length // gcd(loop, length)
if ok:
print(loop if loop > 2 else 1)
``` | instruction | 0 | 99,149 | 14 | 198,298 |
No | output | 1 | 99,149 | 14 | 198,299 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A necklace can be described as a string of links ('-') and pearls ('o'), with the last link or pearl connected to the first one.
<image>
You can remove a link or a pearl and insert it between two other existing links or pearls (or between a link and a pearl) on the necklace. This process can be repeated as many times as you like, but you can't throw away any parts.
Can you make the number of links between every two adjacent pearls equal? Two pearls are considered to be adjacent if there is no other pearl between them.
Note that the final necklace should remain as one circular part of the same length as the initial necklace.
Input
The only line of input contains a string s (3 ≤ |s| ≤ 100), representing the necklace, where a dash '-' represents a link and the lowercase English letter 'o' represents a pearl.
Output
Print "YES" if the links and pearls can be rejoined such that the number of links between adjacent pearls is equal. Otherwise print "NO".
You can print each letter in any case (upper or lower).
Examples
Input
<span class="tex-font-style-tt">-o-o--</span>
Output
YES
Input
<span class="tex-font-style-tt">-o---</span>
Output
YES
Input
<span class="tex-font-style-tt">-o---o-</span>
Output
NO
Input
ooo
Output
YES | instruction | 0 | 99,230 | 14 | 198,460 |
Tags: implementation, math
Correct Solution:
```
s = input()
count1 = count2 = 0
for i in s:
if i =="-":
count1+=1
else:
count2+=1
if count2==0 or count1==0:
print("YES")
else:
if count1%count2==0:
print("YES")
else:
print("NO")
``` | output | 1 | 99,230 | 14 | 198,461 |
Provide a correct Python 3 solution for this coding contest problem.
Niwango-kun has \\(N\\) chickens as his pets. The chickens are identified by numbers \\(1\\) to \\(N\\), and the size of the \\(i\\)-th chicken is a positive integer \\(a_i\\).
\\(N\\) chickens decided to take each other's hand (wing) and form some cycles. The way to make cycles is represented by a permutation \\(p\\) of \\(1, \ldots , N\\). Chicken \\(i\\) takes chicken \\(p_i\\)'s left hand by its right hand. Chickens may take their own hand.
Let us define the cycle containing chicken \\(i\\) as the set consisting of chickens \\(p_i, p_{p_i}, \ldots, p_{\ddots_i} = i\\). It can be proven that after each chicken takes some chicken's hand, the \\(N\\) chickens can be decomposed into cycles.
The beauty \\(f(p)\\) of a way of forming cycles is defined as the product of the size of the smallest chicken in each cycle. Let \\(b_i \ (1 \leq i \leq N)\\) be the sum of \\(f(p)\\) among all possible permutations \\(p\\) for which \\(i\\) cycles are formed in the procedure above.
Find the greatest common divisor of \\(b_1, b_2, \ldots, b_N\\) and print it \\({\rm mod} \ 998244353\\).
Constraints
* \\(1 \leq N \leq 10^5\\)
* \\(1 \leq a_i \leq 10^9\\)
* All numbers given in input are integers
Input
Input is given from Standard Input in the following format:
\(N\)
\(a_1\) \(a_2\) \(\ldots\) \(a_N\)
Output
Print the answer.
Examples
Input
2
4 3
Output
3
Input
4
2 5 2 5
Output
2 | instruction | 0 | 99,325 | 14 | 198,650 |
"Correct Solution:
```
from math import gcd
mod=998244353
N=int(input())
a=list(map(int,input().split()))
a.sort()
ans=1
for i in range(N):
g=gcd(i,a[i])
ans=(ans*g)%mod
print(ans)
``` | output | 1 | 99,325 | 14 | 198,651 |
Provide a correct Python 3 solution for this coding contest problem.
Niwango-kun has \\(N\\) chickens as his pets. The chickens are identified by numbers \\(1\\) to \\(N\\), and the size of the \\(i\\)-th chicken is a positive integer \\(a_i\\).
\\(N\\) chickens decided to take each other's hand (wing) and form some cycles. The way to make cycles is represented by a permutation \\(p\\) of \\(1, \ldots , N\\). Chicken \\(i\\) takes chicken \\(p_i\\)'s left hand by its right hand. Chickens may take their own hand.
Let us define the cycle containing chicken \\(i\\) as the set consisting of chickens \\(p_i, p_{p_i}, \ldots, p_{\ddots_i} = i\\). It can be proven that after each chicken takes some chicken's hand, the \\(N\\) chickens can be decomposed into cycles.
The beauty \\(f(p)\\) of a way of forming cycles is defined as the product of the size of the smallest chicken in each cycle. Let \\(b_i \ (1 \leq i \leq N)\\) be the sum of \\(f(p)\\) among all possible permutations \\(p\\) for which \\(i\\) cycles are formed in the procedure above.
Find the greatest common divisor of \\(b_1, b_2, \ldots, b_N\\) and print it \\({\rm mod} \ 998244353\\).
Constraints
* \\(1 \leq N \leq 10^5\\)
* \\(1 \leq a_i \leq 10^9\\)
* All numbers given in input are integers
Input
Input is given from Standard Input in the following format:
\(N\)
\(a_1\) \(a_2\) \(\ldots\) \(a_N\)
Output
Print the answer.
Examples
Input
2
4 3
Output
3
Input
4
2 5 2 5
Output
2 | instruction | 0 | 99,326 | 14 | 198,652 |
"Correct Solution:
```
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
sys.setrecursionlimit(10 ** 7)
from fractions import gcd
from functools import reduce
MOD = 998244353
N,*A = map(int,read().split())
A.sort()
"""
安い鳥から追加していく。(サイクルの個数 -> 美しさ)が多項式倍で遷移する。
Gaussの補題より、最大個数かけるだけ
"""
answer = reduce(lambda x,y: x*y%MOD, (gcd(i,x) for i,x in enumerate(A)))
print(answer)
``` | output | 1 | 99,326 | 14 | 198,653 |
Provide a correct Python 3 solution for this coding contest problem.
Niwango-kun has \\(N\\) chickens as his pets. The chickens are identified by numbers \\(1\\) to \\(N\\), and the size of the \\(i\\)-th chicken is a positive integer \\(a_i\\).
\\(N\\) chickens decided to take each other's hand (wing) and form some cycles. The way to make cycles is represented by a permutation \\(p\\) of \\(1, \ldots , N\\). Chicken \\(i\\) takes chicken \\(p_i\\)'s left hand by its right hand. Chickens may take their own hand.
Let us define the cycle containing chicken \\(i\\) as the set consisting of chickens \\(p_i, p_{p_i}, \ldots, p_{\ddots_i} = i\\). It can be proven that after each chicken takes some chicken's hand, the \\(N\\) chickens can be decomposed into cycles.
The beauty \\(f(p)\\) of a way of forming cycles is defined as the product of the size of the smallest chicken in each cycle. Let \\(b_i \ (1 \leq i \leq N)\\) be the sum of \\(f(p)\\) among all possible permutations \\(p\\) for which \\(i\\) cycles are formed in the procedure above.
Find the greatest common divisor of \\(b_1, b_2, \ldots, b_N\\) and print it \\({\rm mod} \ 998244353\\).
Constraints
* \\(1 \leq N \leq 10^5\\)
* \\(1 \leq a_i \leq 10^9\\)
* All numbers given in input are integers
Input
Input is given from Standard Input in the following format:
\(N\)
\(a_1\) \(a_2\) \(\ldots\) \(a_N\)
Output
Print the answer.
Examples
Input
2
4 3
Output
3
Input
4
2 5 2 5
Output
2 | instruction | 0 | 99,327 | 14 | 198,654 |
"Correct Solution:
```
N = int(input())
a = list(map(int,input().split()))
Mod = 998244353
def gcd(a, b):
while b:
a, b = b, a % b
return a
a.sort()
ans = 1
for i in range(N):
ans *= gcd(a[i],i)
ans %= Mod
print(ans)
``` | output | 1 | 99,327 | 14 | 198,655 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Niwango-kun has \\(N\\) chickens as his pets. The chickens are identified by numbers \\(1\\) to \\(N\\), and the size of the \\(i\\)-th chicken is a positive integer \\(a_i\\).
\\(N\\) chickens decided to take each other's hand (wing) and form some cycles. The way to make cycles is represented by a permutation \\(p\\) of \\(1, \ldots , N\\). Chicken \\(i\\) takes chicken \\(p_i\\)'s left hand by its right hand. Chickens may take their own hand.
Let us define the cycle containing chicken \\(i\\) as the set consisting of chickens \\(p_i, p_{p_i}, \ldots, p_{\ddots_i} = i\\). It can be proven that after each chicken takes some chicken's hand, the \\(N\\) chickens can be decomposed into cycles.
The beauty \\(f(p)\\) of a way of forming cycles is defined as the product of the size of the smallest chicken in each cycle. Let \\(b_i \ (1 \leq i \leq N)\\) be the sum of \\(f(p)\\) among all possible permutations \\(p\\) for which \\(i\\) cycles are formed in the procedure above.
Find the greatest common divisor of \\(b_1, b_2, \ldots, b_N\\) and print it \\({\rm mod} \ 998244353\\).
Constraints
* \\(1 \leq N \leq 10^5\\)
* \\(1 \leq a_i \leq 10^9\\)
* All numbers given in input are integers
Input
Input is given from Standard Input in the following format:
\(N\)
\(a_1\) \(a_2\) \(\ldots\) \(a_N\)
Output
Print the answer.
Examples
Input
2
4 3
Output
3
Input
4
2 5 2 5
Output
2
Submitted Solution:
```
N=int(input())
L=list(map(int,input().split()))
print(min(L))
``` | instruction | 0 | 99,328 | 14 | 198,656 |
No | output | 1 | 99,328 | 14 | 198,657 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Niwango-kun has \\(N\\) chickens as his pets. The chickens are identified by numbers \\(1\\) to \\(N\\), and the size of the \\(i\\)-th chicken is a positive integer \\(a_i\\).
\\(N\\) chickens decided to take each other's hand (wing) and form some cycles. The way to make cycles is represented by a permutation \\(p\\) of \\(1, \ldots , N\\). Chicken \\(i\\) takes chicken \\(p_i\\)'s left hand by its right hand. Chickens may take their own hand.
Let us define the cycle containing chicken \\(i\\) as the set consisting of chickens \\(p_i, p_{p_i}, \ldots, p_{\ddots_i} = i\\). It can be proven that after each chicken takes some chicken's hand, the \\(N\\) chickens can be decomposed into cycles.
The beauty \\(f(p)\\) of a way of forming cycles is defined as the product of the size of the smallest chicken in each cycle. Let \\(b_i \ (1 \leq i \leq N)\\) be the sum of \\(f(p)\\) among all possible permutations \\(p\\) for which \\(i\\) cycles are formed in the procedure above.
Find the greatest common divisor of \\(b_1, b_2, \ldots, b_N\\) and print it \\({\rm mod} \ 998244353\\).
Constraints
* \\(1 \leq N \leq 10^5\\)
* \\(1 \leq a_i \leq 10^9\\)
* All numbers given in input are integers
Input
Input is given from Standard Input in the following format:
\(N\)
\(a_1\) \(a_2\) \(\ldots\) \(a_N\)
Output
Print the answer.
Examples
Input
2
4 3
Output
3
Input
4
2 5 2 5
Output
2
Submitted Solution:
```
import sys
input = lambda : sys.stdin.readline().rstrip()
sys.setrecursionlimit(max(1000, 10**9))
write = lambda x: sys.stdout.write(x+"\n")
n = int(input())
a = list(map(int, input().split()))
ans = min(a)
print(ans)
``` | instruction | 0 | 99,329 | 14 | 198,658 |
No | output | 1 | 99,329 | 14 | 198,659 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Niwango-kun has \\(N\\) chickens as his pets. The chickens are identified by numbers \\(1\\) to \\(N\\), and the size of the \\(i\\)-th chicken is a positive integer \\(a_i\\).
\\(N\\) chickens decided to take each other's hand (wing) and form some cycles. The way to make cycles is represented by a permutation \\(p\\) of \\(1, \ldots , N\\). Chicken \\(i\\) takes chicken \\(p_i\\)'s left hand by its right hand. Chickens may take their own hand.
Let us define the cycle containing chicken \\(i\\) as the set consisting of chickens \\(p_i, p_{p_i}, \ldots, p_{\ddots_i} = i\\). It can be proven that after each chicken takes some chicken's hand, the \\(N\\) chickens can be decomposed into cycles.
The beauty \\(f(p)\\) of a way of forming cycles is defined as the product of the size of the smallest chicken in each cycle. Let \\(b_i \ (1 \leq i \leq N)\\) be the sum of \\(f(p)\\) among all possible permutations \\(p\\) for which \\(i\\) cycles are formed in the procedure above.
Find the greatest common divisor of \\(b_1, b_2, \ldots, b_N\\) and print it \\({\rm mod} \ 998244353\\).
Constraints
* \\(1 \leq N \leq 10^5\\)
* \\(1 \leq a_i \leq 10^9\\)
* All numbers given in input are integers
Input
Input is given from Standard Input in the following format:
\(N\)
\(a_1\) \(a_2\) \(\ldots\) \(a_N\)
Output
Print the answer.
Examples
Input
2
4 3
Output
3
Input
4
2 5 2 5
Output
2
Submitted Solution:
```
N=int(input())
a=list(map(int,input().split()))
print(min(a))
``` | instruction | 0 | 99,330 | 14 | 198,660 |
No | output | 1 | 99,330 | 14 | 198,661 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Niwango-kun has \\(N\\) chickens as his pets. The chickens are identified by numbers \\(1\\) to \\(N\\), and the size of the \\(i\\)-th chicken is a positive integer \\(a_i\\).
\\(N\\) chickens decided to take each other's hand (wing) and form some cycles. The way to make cycles is represented by a permutation \\(p\\) of \\(1, \ldots , N\\). Chicken \\(i\\) takes chicken \\(p_i\\)'s left hand by its right hand. Chickens may take their own hand.
Let us define the cycle containing chicken \\(i\\) as the set consisting of chickens \\(p_i, p_{p_i}, \ldots, p_{\ddots_i} = i\\). It can be proven that after each chicken takes some chicken's hand, the \\(N\\) chickens can be decomposed into cycles.
The beauty \\(f(p)\\) of a way of forming cycles is defined as the product of the size of the smallest chicken in each cycle. Let \\(b_i \ (1 \leq i \leq N)\\) be the sum of \\(f(p)\\) among all possible permutations \\(p\\) for which \\(i\\) cycles are formed in the procedure above.
Find the greatest common divisor of \\(b_1, b_2, \ldots, b_N\\) and print it \\({\rm mod} \ 998244353\\).
Constraints
* \\(1 \leq N \leq 10^5\\)
* \\(1 \leq a_i \leq 10^9\\)
* All numbers given in input are integers
Input
Input is given from Standard Input in the following format:
\(N\)
\(a_1\) \(a_2\) \(\ldots\) \(a_N\)
Output
Print the answer.
Examples
Input
2
4 3
Output
3
Input
4
2 5 2 5
Output
2
Submitted Solution:
```
import numpy as np
def main():
n = int(input())
a = list(map(int, input().split()))
m = np.mean(a)
r = a[np.argmin((a-m)**2)]
print(r)
main()
``` | instruction | 0 | 99,331 | 14 | 198,662 |
No | output | 1 | 99,331 | 14 | 198,663 |
Provide a correct Python 3 solution for this coding contest problem.
$N$ sages are sitting around a round table with $N$ seats. Each sage holds chopsticks with his dominant hand to eat his dinner. The following happens in this situation.
* If sage $i$ is right-handed and a left-handed sage sits on his right, a level of frustration $w_i$ occurs to him. A right-handed sage on his right does not cause such frustration at all.
* If sage $i$ is left-handed and a right-handed sage sits on his left, a level of frustration $w_i$ occurs to him. A left-handed sage on his left does not cause such frustration at all.
You wish you could minimize the total amount of frustration by clever sitting order arrangement.
Given the number of sages with his dominant hand information, make a program to evaluate the minimum frustration achievable.
Input
The input is given in the following format.
$N$
$a_1$ $a_2$ $...$ $a_N$
$w_1$ $w_2$ $...$ $w_N$
The first line provides the number of sages $N$ ($3 \leq N \leq 10$). The second line provides an array of integers $a_i$ (0 or 1) which indicate if the $i$-th sage is right-handed (0) or left-handed (1). The third line provides an array of integers $w_i$ ($1 \leq w_i \leq 1000$) which indicate the level of frustration the $i$-th sage bears.
Output
Output the minimum total frustration the sages bear.
Examples
Input
5
1 0 0 1 0
2 3 5 1 2
Output
3
Input
3
0 0 0
1 2 3
Output
0 | instruction | 0 | 99,373 | 14 | 198,746 |
"Correct Solution:
```
N = int(input())
a = [int(i) for i in input().split()]
w = [int(i) for i in input().split()]
rightMin = 1001
leftMin = 1001
for i in range(N):
if a[i] == 0:
rightMin = min(rightMin, w[i])
else:
leftMin = min(leftMin, w[i])
if rightMin == 1001 or leftMin == 1001:
print(0)
else:
print(rightMin + leftMin)
``` | output | 1 | 99,373 | 14 | 198,747 |
Provide a correct Python 3 solution for this coding contest problem.
$N$ sages are sitting around a round table with $N$ seats. Each sage holds chopsticks with his dominant hand to eat his dinner. The following happens in this situation.
* If sage $i$ is right-handed and a left-handed sage sits on his right, a level of frustration $w_i$ occurs to him. A right-handed sage on his right does not cause such frustration at all.
* If sage $i$ is left-handed and a right-handed sage sits on his left, a level of frustration $w_i$ occurs to him. A left-handed sage on his left does not cause such frustration at all.
You wish you could minimize the total amount of frustration by clever sitting order arrangement.
Given the number of sages with his dominant hand information, make a program to evaluate the minimum frustration achievable.
Input
The input is given in the following format.
$N$
$a_1$ $a_2$ $...$ $a_N$
$w_1$ $w_2$ $...$ $w_N$
The first line provides the number of sages $N$ ($3 \leq N \leq 10$). The second line provides an array of integers $a_i$ (0 or 1) which indicate if the $i$-th sage is right-handed (0) or left-handed (1). The third line provides an array of integers $w_i$ ($1 \leq w_i \leq 1000$) which indicate the level of frustration the $i$-th sage bears.
Output
Output the minimum total frustration the sages bear.
Examples
Input
5
1 0 0 1 0
2 3 5 1 2
Output
3
Input
3
0 0 0
1 2 3
Output
0 | instruction | 0 | 99,374 | 14 | 198,748 |
"Correct Solution:
```
n = int(input())
a = list(map(int,input().split()))
w = list(map(int,input().split()))
INF = 10**9
x = [INF]*2
for i in range(n):
x[a[i]] = min(x[a[i]], w[i])
ans = sum(x)
if ans >= INF:
ans = 0
print(ans)
``` | output | 1 | 99,374 | 14 | 198,749 |
Provide a correct Python 3 solution for this coding contest problem.
$N$ sages are sitting around a round table with $N$ seats. Each sage holds chopsticks with his dominant hand to eat his dinner. The following happens in this situation.
* If sage $i$ is right-handed and a left-handed sage sits on his right, a level of frustration $w_i$ occurs to him. A right-handed sage on his right does not cause such frustration at all.
* If sage $i$ is left-handed and a right-handed sage sits on his left, a level of frustration $w_i$ occurs to him. A left-handed sage on his left does not cause such frustration at all.
You wish you could minimize the total amount of frustration by clever sitting order arrangement.
Given the number of sages with his dominant hand information, make a program to evaluate the minimum frustration achievable.
Input
The input is given in the following format.
$N$
$a_1$ $a_2$ $...$ $a_N$
$w_1$ $w_2$ $...$ $w_N$
The first line provides the number of sages $N$ ($3 \leq N \leq 10$). The second line provides an array of integers $a_i$ (0 or 1) which indicate if the $i$-th sage is right-handed (0) or left-handed (1). The third line provides an array of integers $w_i$ ($1 \leq w_i \leq 1000$) which indicate the level of frustration the $i$-th sage bears.
Output
Output the minimum total frustration the sages bear.
Examples
Input
5
1 0 0 1 0
2 3 5 1 2
Output
3
Input
3
0 0 0
1 2 3
Output
0 | instruction | 0 | 99,375 | 14 | 198,750 |
"Correct Solution:
```
N = input()
a = [int(x) for x in input().split()]
w = [int(x) for x in input().split()]
R = [w for a, w in zip(a, w) if a == 0]
L = [w for a, w in zip(a, w) if a == 1]
if R and L:
print(min(R) + min(L))
else:
print(0)
``` | output | 1 | 99,375 | 14 | 198,751 |
Provide a correct Python 3 solution for this coding contest problem.
$N$ sages are sitting around a round table with $N$ seats. Each sage holds chopsticks with his dominant hand to eat his dinner. The following happens in this situation.
* If sage $i$ is right-handed and a left-handed sage sits on his right, a level of frustration $w_i$ occurs to him. A right-handed sage on his right does not cause such frustration at all.
* If sage $i$ is left-handed and a right-handed sage sits on his left, a level of frustration $w_i$ occurs to him. A left-handed sage on his left does not cause such frustration at all.
You wish you could minimize the total amount of frustration by clever sitting order arrangement.
Given the number of sages with his dominant hand information, make a program to evaluate the minimum frustration achievable.
Input
The input is given in the following format.
$N$
$a_1$ $a_2$ $...$ $a_N$
$w_1$ $w_2$ $...$ $w_N$
The first line provides the number of sages $N$ ($3 \leq N \leq 10$). The second line provides an array of integers $a_i$ (0 or 1) which indicate if the $i$-th sage is right-handed (0) or left-handed (1). The third line provides an array of integers $w_i$ ($1 \leq w_i \leq 1000$) which indicate the level of frustration the $i$-th sage bears.
Output
Output the minimum total frustration the sages bear.
Examples
Input
5
1 0 0 1 0
2 3 5 1 2
Output
3
Input
3
0 0 0
1 2 3
Output
0 | instruction | 0 | 99,376 | 14 | 198,752 |
"Correct Solution:
```
n = int(input())
alst = list(map(int, input().split()))
wlst = list(map(int, input().split()))
right = [w for a, w in zip(alst, wlst) if a == 0]
left = [w for a, w in zip(alst, wlst) if a == 1]
if right and left:
print(min(right) + min(left))
else:
print(0)
``` | output | 1 | 99,376 | 14 | 198,753 |
Provide a correct Python 3 solution for this coding contest problem.
$N$ sages are sitting around a round table with $N$ seats. Each sage holds chopsticks with his dominant hand to eat his dinner. The following happens in this situation.
* If sage $i$ is right-handed and a left-handed sage sits on his right, a level of frustration $w_i$ occurs to him. A right-handed sage on his right does not cause such frustration at all.
* If sage $i$ is left-handed and a right-handed sage sits on his left, a level of frustration $w_i$ occurs to him. A left-handed sage on his left does not cause such frustration at all.
You wish you could minimize the total amount of frustration by clever sitting order arrangement.
Given the number of sages with his dominant hand information, make a program to evaluate the minimum frustration achievable.
Input
The input is given in the following format.
$N$
$a_1$ $a_2$ $...$ $a_N$
$w_1$ $w_2$ $...$ $w_N$
The first line provides the number of sages $N$ ($3 \leq N \leq 10$). The second line provides an array of integers $a_i$ (0 or 1) which indicate if the $i$-th sage is right-handed (0) or left-handed (1). The third line provides an array of integers $w_i$ ($1 \leq w_i \leq 1000$) which indicate the level of frustration the $i$-th sage bears.
Output
Output the minimum total frustration the sages bear.
Examples
Input
5
1 0 0 1 0
2 3 5 1 2
Output
3
Input
3
0 0 0
1 2 3
Output
0 | instruction | 0 | 99,377 | 14 | 198,754 |
"Correct Solution:
```
N = int(input())
*A, = map(int, input().split())
*W, = map(int, input().split())
INF = 10**9
l = r = INF
for a, w in zip(A, W):
if a:
r = min(w, r)
else:
l = min(w, l)
if l == INF or r == INF:
print(0)
else:
print(l+r)
``` | output | 1 | 99,377 | 14 | 198,755 |
Provide a correct Python 3 solution for this coding contest problem.
$N$ sages are sitting around a round table with $N$ seats. Each sage holds chopsticks with his dominant hand to eat his dinner. The following happens in this situation.
* If sage $i$ is right-handed and a left-handed sage sits on his right, a level of frustration $w_i$ occurs to him. A right-handed sage on his right does not cause such frustration at all.
* If sage $i$ is left-handed and a right-handed sage sits on his left, a level of frustration $w_i$ occurs to him. A left-handed sage on his left does not cause such frustration at all.
You wish you could minimize the total amount of frustration by clever sitting order arrangement.
Given the number of sages with his dominant hand information, make a program to evaluate the minimum frustration achievable.
Input
The input is given in the following format.
$N$
$a_1$ $a_2$ $...$ $a_N$
$w_1$ $w_2$ $...$ $w_N$
The first line provides the number of sages $N$ ($3 \leq N \leq 10$). The second line provides an array of integers $a_i$ (0 or 1) which indicate if the $i$-th sage is right-handed (0) or left-handed (1). The third line provides an array of integers $w_i$ ($1 \leq w_i \leq 1000$) which indicate the level of frustration the $i$-th sage bears.
Output
Output the minimum total frustration the sages bear.
Examples
Input
5
1 0 0 1 0
2 3 5 1 2
Output
3
Input
3
0 0 0
1 2 3
Output
0 | instruction | 0 | 99,378 | 14 | 198,756 |
"Correct Solution:
```
n = int(input())
a = [int(_) for _ in input().split()]
w = [int(_) for _ in input().split()]
if sum(a) == n or sum(a) == 0:
print(0)
quit()
lia = []
lib = []
for i in range(n):
if a[i] == 0:
lia.append(w[i])
else:
lib.append(w[i])
print(min(lia)+min(lib))
``` | output | 1 | 99,378 | 14 | 198,757 |
Provide a correct Python 3 solution for this coding contest problem.
$N$ sages are sitting around a round table with $N$ seats. Each sage holds chopsticks with his dominant hand to eat his dinner. The following happens in this situation.
* If sage $i$ is right-handed and a left-handed sage sits on his right, a level of frustration $w_i$ occurs to him. A right-handed sage on his right does not cause such frustration at all.
* If sage $i$ is left-handed and a right-handed sage sits on his left, a level of frustration $w_i$ occurs to him. A left-handed sage on his left does not cause such frustration at all.
You wish you could minimize the total amount of frustration by clever sitting order arrangement.
Given the number of sages with his dominant hand information, make a program to evaluate the minimum frustration achievable.
Input
The input is given in the following format.
$N$
$a_1$ $a_2$ $...$ $a_N$
$w_1$ $w_2$ $...$ $w_N$
The first line provides the number of sages $N$ ($3 \leq N \leq 10$). The second line provides an array of integers $a_i$ (0 or 1) which indicate if the $i$-th sage is right-handed (0) or left-handed (1). The third line provides an array of integers $w_i$ ($1 \leq w_i \leq 1000$) which indicate the level of frustration the $i$-th sage bears.
Output
Output the minimum total frustration the sages bear.
Examples
Input
5
1 0 0 1 0
2 3 5 1 2
Output
3
Input
3
0 0 0
1 2 3
Output
0 | instruction | 0 | 99,379 | 14 | 198,758 |
"Correct Solution:
```
n = int(input())
aw = [[0,0] for i in range(n)]
for i, a in enumerate(map(int, input().split())): aw[i][0] = a
for i, w in enumerate(map(int, input().split())): aw[i][1] = w
migimin = 1001
hidarimin = 1001
for a,w in aw:
if a: hidarimin = min(hidarimin, w)
else: migimin = min(migimin, w)
if hidarimin>1000 or migimin>1000: print(0)
else: print(hidarimin+migimin)
``` | output | 1 | 99,379 | 14 | 198,759 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Jack has become a soldier now. Unfortunately, he has trouble with the drill. Instead of marching beginning with the left foot and then changing legs with each step, as ordered, he keeps repeating a sequence of steps, in which he sometimes makes the wrong steps or — horror of horrors! — stops for a while. For example, if Jack uses the sequence 'right, left, break', when the sergeant yells: 'Left! Right! Left! Right! Left! Right!', Jack first makes a step with the right foot, then one with the left foot, then he is confused and stops for a moment, then again - this time according to the order - starts with the right foot, then uses the left foot, then - to the sergeant's irritation - he stops to catch his breath, to incorrectly start with the right foot again... Marching this way, Jack will make the step that he is supposed to in the given moment in only one third of cases.
When the officers convinced him he should do something about it, Jack decided to modify the basic sequence of steps that he repeats. However, in order not to get too tired, he has decided that the only thing he'll do is adding any number of breaks in any positions of the original sequence (a break corresponds to stopping for the duration of one step). Of course, Jack can't make a step on the same foot twice in a row, if there is no pause between these steps. It is, however, not impossible that the sequence of steps he used so far is incorrect (it would explain a lot, actually).
Help Private Jack! Given the sequence of steps he keeps repeating, calculate the maximal percentage of time that he can spend marching correctly after adding some breaks to his scheme.
Input
The first line of input contains a sequence consisting only of characters 'L', 'R' and 'X', where 'L' corresponds to a step with the left foot, 'R' — with the right foot, and 'X' — to a break. The length of the sequence will not exceed 106.
Output
Output the maximum percentage of time that Jack can spend marching correctly, rounded down to exactly six digits after the decimal point.
Examples
Input
X
Output
0.000000
Input
LXRR
Output
50.000000
Note
In the second example, if we add two breaks to receive LXXRXR, Jack will march: LXXRXRLXXRXRL... instead of LRLRLRLRLRLRL... and will make the correct step in half the cases. If we didn't add any breaks, the sequence would be incorrect — Jack can't step on his right foot twice in a row. | instruction | 0 | 99,567 | 14 | 199,134 |
Tags: binary search, dp, greedy
Correct Solution:
```
def solve():
original = input()
temp = [original[0]]
for i in range(1, len(original)):
if original[i] == original[i - 1] != 'X':
temp.append('X')
temp.append(original[i])
augmented = ''.join(temp)
answer = 0
if augmented[0] == augmented[-1] != 'X':
answer = max(rate(augmented + 'X'), rate('X' + augmented))
else:
answer = rate(augmented)
print('%d.%06d' % (answer / 1000000, answer % 1000000))
def rate(seq):
correct, total, unknown, indicator = 0, 0, 0, 0
left_step = True
for action in seq:
if action == 'X':
total += 1
left_step = not left_step
else:
if left_step and action == 'L' or not left_step and action == 'R':
correct += 1
total += 1
indicator = 0
left_step = not left_step
else:
correct += 1
total += 2
unknown += indicator
indicator = 1 - indicator
if total % 2 == 1:
total += 1
unknown += indicator
if correct * 2 > total:
correct -= unknown
total -= unknown * 2
return correct * 100000000 // total
solve()
``` | output | 1 | 99,567 | 14 | 199,135 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Jack has become a soldier now. Unfortunately, he has trouble with the drill. Instead of marching beginning with the left foot and then changing legs with each step, as ordered, he keeps repeating a sequence of steps, in which he sometimes makes the wrong steps or — horror of horrors! — stops for a while. For example, if Jack uses the sequence 'right, left, break', when the sergeant yells: 'Left! Right! Left! Right! Left! Right!', Jack first makes a step with the right foot, then one with the left foot, then he is confused and stops for a moment, then again - this time according to the order - starts with the right foot, then uses the left foot, then - to the sergeant's irritation - he stops to catch his breath, to incorrectly start with the right foot again... Marching this way, Jack will make the step that he is supposed to in the given moment in only one third of cases.
When the officers convinced him he should do something about it, Jack decided to modify the basic sequence of steps that he repeats. However, in order not to get too tired, he has decided that the only thing he'll do is adding any number of breaks in any positions of the original sequence (a break corresponds to stopping for the duration of one step). Of course, Jack can't make a step on the same foot twice in a row, if there is no pause between these steps. It is, however, not impossible that the sequence of steps he used so far is incorrect (it would explain a lot, actually).
Help Private Jack! Given the sequence of steps he keeps repeating, calculate the maximal percentage of time that he can spend marching correctly after adding some breaks to his scheme.
Input
The first line of input contains a sequence consisting only of characters 'L', 'R' and 'X', where 'L' corresponds to a step with the left foot, 'R' — with the right foot, and 'X' — to a break. The length of the sequence will not exceed 106.
Output
Output the maximum percentage of time that Jack can spend marching correctly, rounded down to exactly six digits after the decimal point.
Examples
Input
X
Output
0.000000
Input
LXRR
Output
50.000000
Note
In the second example, if we add two breaks to receive LXXRXR, Jack will march: LXXRXRLXXRXRL... instead of LRLRLRLRLRLRL... and will make the correct step in half the cases. If we didn't add any breaks, the sequence would be incorrect — Jack can't step on his right foot twice in a row. | instruction | 0 | 99,568 | 14 | 199,136 |
Tags: binary search, dp, greedy
Correct Solution:
```
__author__ = 'Darren'
def solve():
original = input()
temp = [original[0]]
for i in range(1, len(original)):
if original[i] == original[i-1] != 'X':
temp.append('X')
temp.append(original[i])
augmented = ''.join(temp)
answer = 0
if augmented[0] == augmented[-1] != 'X':
answer = max(rate(augmented+'X'), rate('X'+augmented))
else:
answer = rate(augmented)
print('%d.%06d' % (answer / 1000000, answer % 1000000))
def rate(seq):
correct, total, unknown, indicator = 0, 0, 0, 0
left_step = True
for action in seq:
if action == 'X':
total += 1
left_step = not left_step
else:
if left_step and action == 'L' or not left_step and action == 'R':
correct += 1
total += 1
indicator = 0
left_step = not left_step
else:
correct += 1
total += 2
unknown += indicator
indicator = 1 - indicator
if total % 2 == 1:
total += 1
unknown += indicator
if correct * 2 > total:
correct -= unknown
total -= unknown * 2
return correct * 100000000 // total
if __name__ == '__main__':
solve()
``` | output | 1 | 99,568 | 14 | 199,137 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easy version of the problem. The difference is constraints on the number of wise men and the time limit. You can make hacks only if all versions of this task are solved.
n wise men live in a beautiful city. Some of them know each other.
For each of the n! possible permutations p_1, p_2, …, p_n of the wise men, let's generate a binary string of length n-1: for each 1 ≤ i < n set s_i=1 if p_i and p_{i+1} know each other, and s_i=0 otherwise.
For all possible 2^{n-1} binary strings, find the number of permutations that produce this binary string.
Input
The first line of input contains one integer n (2 ≤ n ≤ 14) — the number of wise men in the city.
The next n lines contain a binary string of length n each, such that the j-th character of the i-th string is equal to '1' if wise man i knows wise man j, and equals '0' otherwise.
It is guaranteed that if the i-th man knows the j-th man, then the j-th man knows i-th man and no man knows himself.
Output
Print 2^{n-1} space-separated integers. For each 0 ≤ x < 2^{n-1}:
* Let's consider a string s of length n-1, such that s_i = ⌊ \frac{x}{2^{i-1}} ⌋ mod 2 for all 1 ≤ i ≤ n - 1.
* The (x+1)-th number should be equal to the required answer for s.
Examples
Input
3
011
101
110
Output
0 0 0 6
Input
4
0101
1000
0001
1010
Output
2 2 6 2 2 6 2 2
Note
In the first test, each wise man knows each other, so every permutation will produce the string 11.
In the second test:
* If p = \{1, 2, 3, 4\}, the produced string is 101, because wise men 1 and 2 know each other, 2 and 3 don't know each other, and 3 and 4 know each other;
* If p = \{4, 1, 2, 3\}, the produced string is 110, because wise men 1 and 4 know each other, 1 and 2 know each other and 2, and 3 don't know each other;
* If p = \{1, 3, 2, 4\}, the produced string is 000, because wise men 1 and 3 don't know each other, 3 and 2 don't know each other, and 2 and 4 don't know each other. | instruction | 0 | 99,592 | 14 | 199,184 |
Tags: bitmasks, brute force, dp, meet-in-the-middle
Correct Solution:
```
from sys import stdout
n = int(input())
class Person:
num = n - 1
def __init__(self, rel):
self.relationship = int(rel, 2)
def __getitem__(self, k):
return (self.relationship >> Person.num - k) & 1
rel = [Person(input()) for _ in range(n)]
dp = [[0] * n for _ in range(1 << n)]
for people in range(1, 1 << n):
ones = [i for i in range(n) if people & (1 << i)]
# print(f'ones: {ones}')
one_num = len(ones)
if one_num == 1:
dp[people][ones[0]] = [1]
continue
for i in ones:
dp[people][i] = [0] * (1 << one_num - 1)
pre_people = people ^ (1 << i)
for j in ones:
if j == i:
continue
for pre_s, times in enumerate(dp[pre_people][j]):
s = pre_s | (rel[j][i] << one_num - 2)
# print(f'dp[{people}][{i}][{s}]: {dp[people][i][s]}')
dp[people][i][s] += times
people = (1 << n) - 1
for s in range(1 << (n-1)):
ans = 0
for i in range(n):
ans += dp[people][i][s]
print(ans, end=' ')
``` | output | 1 | 99,592 | 14 | 199,185 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As a tradition, every year before IOI all the members of Natalia Fan Club are invited to Malek Dance Club to have a fun night together. Malek Dance Club has 2n members and coincidentally Natalia Fan Club also has 2n members. Each member of MDC is assigned a unique id i from 0 to 2n - 1. The same holds for each member of NFC.
One of the parts of this tradition is one by one dance, where each member of MDC dances with a member of NFC. A dance pair is a pair of numbers (a, b) such that member a from MDC dances with member b from NFC.
The complexity of a pairs' assignment is the number of pairs of dancing pairs (a, b) and (c, d) such that a < c and b > d.
You are given a binary number of length n named x. We know that member i from MDC dances with member <image> from NFC. Your task is to calculate the complexity of this assignment modulo 1000000007 (109 + 7).
Expression <image> denotes applying «XOR» to numbers x and y. This operation exists in all modern programming languages, for example, in C++ and Java it denotes as «^», in Pascal — «xor».
Input
The first line of input contains a binary number x of lenght n, (1 ≤ n ≤ 100).
This number may contain leading zeros.
Output
Print the complexity of the given dance assignent modulo 1000000007 (109 + 7).
Examples
Input
11
Output
6
Input
01
Output
2
Input
1
Output
1 | instruction | 0 | 99,730 | 14 | 199,460 |
Tags: combinatorics, math
Correct Solution:
```
x = input()
n = len(x)
mod = 10 ** 9 + 7
ans = 0
for i in range(n):
if x[i] == '1':
pref = pow(2, i, mod)
suf = pow(2, n - i - 1, mod) ** 2
ans += (pref * suf) % mod
print(ans % mod)
``` | output | 1 | 99,730 | 14 | 199,461 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As a tradition, every year before IOI all the members of Natalia Fan Club are invited to Malek Dance Club to have a fun night together. Malek Dance Club has 2n members and coincidentally Natalia Fan Club also has 2n members. Each member of MDC is assigned a unique id i from 0 to 2n - 1. The same holds for each member of NFC.
One of the parts of this tradition is one by one dance, where each member of MDC dances with a member of NFC. A dance pair is a pair of numbers (a, b) such that member a from MDC dances with member b from NFC.
The complexity of a pairs' assignment is the number of pairs of dancing pairs (a, b) and (c, d) such that a < c and b > d.
You are given a binary number of length n named x. We know that member i from MDC dances with member <image> from NFC. Your task is to calculate the complexity of this assignment modulo 1000000007 (109 + 7).
Expression <image> denotes applying «XOR» to numbers x and y. This operation exists in all modern programming languages, for example, in C++ and Java it denotes as «^», in Pascal — «xor».
Input
The first line of input contains a binary number x of lenght n, (1 ≤ n ≤ 100).
This number may contain leading zeros.
Output
Print the complexity of the given dance assignent modulo 1000000007 (109 + 7).
Examples
Input
11
Output
6
Input
01
Output
2
Input
1
Output
1 | instruction | 0 | 99,731 | 14 | 199,462 |
Tags: combinatorics, math
Correct Solution:
```
#!/usr/local/bin/python3.3 -tt
import sys
if __name__ == '__main__':
for l in sys.stdin:
s = l.strip()
break
n = len(s)
d = 2 ** (n - 1)
print(d * int(s, 2) % 1000000007)
``` | output | 1 | 99,731 | 14 | 199,463 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As a tradition, every year before IOI all the members of Natalia Fan Club are invited to Malek Dance Club to have a fun night together. Malek Dance Club has 2n members and coincidentally Natalia Fan Club also has 2n members. Each member of MDC is assigned a unique id i from 0 to 2n - 1. The same holds for each member of NFC.
One of the parts of this tradition is one by one dance, where each member of MDC dances with a member of NFC. A dance pair is a pair of numbers (a, b) such that member a from MDC dances with member b from NFC.
The complexity of a pairs' assignment is the number of pairs of dancing pairs (a, b) and (c, d) such that a < c and b > d.
You are given a binary number of length n named x. We know that member i from MDC dances with member <image> from NFC. Your task is to calculate the complexity of this assignment modulo 1000000007 (109 + 7).
Expression <image> denotes applying «XOR» to numbers x and y. This operation exists in all modern programming languages, for example, in C++ and Java it denotes as «^», in Pascal — «xor».
Input
The first line of input contains a binary number x of lenght n, (1 ≤ n ≤ 100).
This number may contain leading zeros.
Output
Print the complexity of the given dance assignent modulo 1000000007 (109 + 7).
Examples
Input
11
Output
6
Input
01
Output
2
Input
1
Output
1 | instruction | 0 | 99,732 | 14 | 199,464 |
Tags: combinatorics, math
Correct Solution:
```
s = input()
s = s[::-1]
ans = 0
for i in range(len(s)):
if s[i] == '1':
ans += ((2 ** i) ** 2) * (2**(len(s) - i - 1))
print(ans % (10**9+7))
``` | output | 1 | 99,732 | 14 | 199,465 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As a tradition, every year before IOI all the members of Natalia Fan Club are invited to Malek Dance Club to have a fun night together. Malek Dance Club has 2n members and coincidentally Natalia Fan Club also has 2n members. Each member of MDC is assigned a unique id i from 0 to 2n - 1. The same holds for each member of NFC.
One of the parts of this tradition is one by one dance, where each member of MDC dances with a member of NFC. A dance pair is a pair of numbers (a, b) such that member a from MDC dances with member b from NFC.
The complexity of a pairs' assignment is the number of pairs of dancing pairs (a, b) and (c, d) such that a < c and b > d.
You are given a binary number of length n named x. We know that member i from MDC dances with member <image> from NFC. Your task is to calculate the complexity of this assignment modulo 1000000007 (109 + 7).
Expression <image> denotes applying «XOR» to numbers x and y. This operation exists in all modern programming languages, for example, in C++ and Java it denotes as «^», in Pascal — «xor».
Input
The first line of input contains a binary number x of lenght n, (1 ≤ n ≤ 100).
This number may contain leading zeros.
Output
Print the complexity of the given dance assignent modulo 1000000007 (109 + 7).
Examples
Input
11
Output
6
Input
01
Output
2
Input
1
Output
1 | instruction | 0 | 99,733 | 14 | 199,466 |
Tags: combinatorics, math
Correct Solution:
```
p = 1000000007
s = input()
print(int(s,2) * pow(2, len(s) - 1) % p)
# Made By Mostafa_Khaled
``` | output | 1 | 99,733 | 14 | 199,467 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As a tradition, every year before IOI all the members of Natalia Fan Club are invited to Malek Dance Club to have a fun night together. Malek Dance Club has 2n members and coincidentally Natalia Fan Club also has 2n members. Each member of MDC is assigned a unique id i from 0 to 2n - 1. The same holds for each member of NFC.
One of the parts of this tradition is one by one dance, where each member of MDC dances with a member of NFC. A dance pair is a pair of numbers (a, b) such that member a from MDC dances with member b from NFC.
The complexity of a pairs' assignment is the number of pairs of dancing pairs (a, b) and (c, d) such that a < c and b > d.
You are given a binary number of length n named x. We know that member i from MDC dances with member <image> from NFC. Your task is to calculate the complexity of this assignment modulo 1000000007 (109 + 7).
Expression <image> denotes applying «XOR» to numbers x and y. This operation exists in all modern programming languages, for example, in C++ and Java it denotes as «^», in Pascal — «xor».
Input
The first line of input contains a binary number x of lenght n, (1 ≤ n ≤ 100).
This number may contain leading zeros.
Output
Print the complexity of the given dance assignent modulo 1000000007 (109 + 7).
Examples
Input
11
Output
6
Input
01
Output
2
Input
1
Output
1 | instruction | 0 | 99,734 | 14 | 199,468 |
Tags: combinatorics, math
Correct Solution:
```
n = input().strip()
s = len(n)
k = int(n,2)
start = 4 **( s-1)
zib = 2**(s-1)
step = 2**(s-1)
print((start+(k-zib)*step)%(10**9+7))
``` | output | 1 | 99,734 | 14 | 199,469 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As a tradition, every year before IOI all the members of Natalia Fan Club are invited to Malek Dance Club to have a fun night together. Malek Dance Club has 2n members and coincidentally Natalia Fan Club also has 2n members. Each member of MDC is assigned a unique id i from 0 to 2n - 1. The same holds for each member of NFC.
One of the parts of this tradition is one by one dance, where each member of MDC dances with a member of NFC. A dance pair is a pair of numbers (a, b) such that member a from MDC dances with member b from NFC.
The complexity of a pairs' assignment is the number of pairs of dancing pairs (a, b) and (c, d) such that a < c and b > d.
You are given a binary number of length n named x. We know that member i from MDC dances with member <image> from NFC. Your task is to calculate the complexity of this assignment modulo 1000000007 (109 + 7).
Expression <image> denotes applying «XOR» to numbers x and y. This operation exists in all modern programming languages, for example, in C++ and Java it denotes as «^», in Pascal — «xor».
Input
The first line of input contains a binary number x of lenght n, (1 ≤ n ≤ 100).
This number may contain leading zeros.
Output
Print the complexity of the given dance assignent modulo 1000000007 (109 + 7).
Examples
Input
11
Output
6
Input
01
Output
2
Input
1
Output
1 | instruction | 0 | 99,735 | 14 | 199,470 |
Tags: combinatorics, math
Correct Solution:
```
MOD = int(1e9 + 7)
x = input()[::-1]
n = len(x)
res = 0
for i, t in enumerate(x):
if t == '1':
res = (res + (1 << (n - 1 + i))) % MOD
print(res)
``` | output | 1 | 99,735 | 14 | 199,471 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As a tradition, every year before IOI all the members of Natalia Fan Club are invited to Malek Dance Club to have a fun night together. Malek Dance Club has 2n members and coincidentally Natalia Fan Club also has 2n members. Each member of MDC is assigned a unique id i from 0 to 2n - 1. The same holds for each member of NFC.
One of the parts of this tradition is one by one dance, where each member of MDC dances with a member of NFC. A dance pair is a pair of numbers (a, b) such that member a from MDC dances with member b from NFC.
The complexity of a pairs' assignment is the number of pairs of dancing pairs (a, b) and (c, d) such that a < c and b > d.
You are given a binary number of length n named x. We know that member i from MDC dances with member <image> from NFC. Your task is to calculate the complexity of this assignment modulo 1000000007 (109 + 7).
Expression <image> denotes applying «XOR» to numbers x and y. This operation exists in all modern programming languages, for example, in C++ and Java it denotes as «^», in Pascal — «xor».
Input
The first line of input contains a binary number x of lenght n, (1 ≤ n ≤ 100).
This number may contain leading zeros.
Output
Print the complexity of the given dance assignent modulo 1000000007 (109 + 7).
Examples
Input
11
Output
6
Input
01
Output
2
Input
1
Output
1 | instruction | 0 | 99,736 | 14 | 199,472 |
Tags: combinatorics, math
Correct Solution:
```
MOD = 1000000007
ans = 0
x = input().strip()
n = len(x)
for i, v in enumerate(x):
if v == "1":
ans += pow(2, i, MOD) * pow(2, (n - i - 1) * 2, MOD) % MOD
ans %= MOD
print(ans)
``` | output | 1 | 99,736 | 14 | 199,473 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As a tradition, every year before IOI all the members of Natalia Fan Club are invited to Malek Dance Club to have a fun night together. Malek Dance Club has 2n members and coincidentally Natalia Fan Club also has 2n members. Each member of MDC is assigned a unique id i from 0 to 2n - 1. The same holds for each member of NFC.
One of the parts of this tradition is one by one dance, where each member of MDC dances with a member of NFC. A dance pair is a pair of numbers (a, b) such that member a from MDC dances with member b from NFC.
The complexity of a pairs' assignment is the number of pairs of dancing pairs (a, b) and (c, d) such that a < c and b > d.
You are given a binary number of length n named x. We know that member i from MDC dances with member <image> from NFC. Your task is to calculate the complexity of this assignment modulo 1000000007 (109 + 7).
Expression <image> denotes applying «XOR» to numbers x and y. This operation exists in all modern programming languages, for example, in C++ and Java it denotes as «^», in Pascal — «xor».
Input
The first line of input contains a binary number x of lenght n, (1 ≤ n ≤ 100).
This number may contain leading zeros.
Output
Print the complexity of the given dance assignent modulo 1000000007 (109 + 7).
Examples
Input
11
Output
6
Input
01
Output
2
Input
1
Output
1
Submitted Solution:
```
MOD = int(10e9 + 7)
x = input()[::-1]
n = len(x)
res = 0
for i, t in enumerate(x):
if t == '1':
res = (res + (1 << (n - 1 + i))) % MOD
print(res)
``` | instruction | 0 | 99,737 | 14 | 199,474 |
No | output | 1 | 99,737 | 14 | 199,475 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As a tradition, every year before IOI all the members of Natalia Fan Club are invited to Malek Dance Club to have a fun night together. Malek Dance Club has 2n members and coincidentally Natalia Fan Club also has 2n members. Each member of MDC is assigned a unique id i from 0 to 2n - 1. The same holds for each member of NFC.
One of the parts of this tradition is one by one dance, where each member of MDC dances with a member of NFC. A dance pair is a pair of numbers (a, b) such that member a from MDC dances with member b from NFC.
The complexity of a pairs' assignment is the number of pairs of dancing pairs (a, b) and (c, d) such that a < c and b > d.
You are given a binary number of length n named x. We know that member i from MDC dances with member <image> from NFC. Your task is to calculate the complexity of this assignment modulo 1000000007 (109 + 7).
Expression <image> denotes applying «XOR» to numbers x and y. This operation exists in all modern programming languages, for example, in C++ and Java it denotes as «^», in Pascal — «xor».
Input
The first line of input contains a binary number x of lenght n, (1 ≤ n ≤ 100).
This number may contain leading zeros.
Output
Print the complexity of the given dance assignent modulo 1000000007 (109 + 7).
Examples
Input
11
Output
6
Input
01
Output
2
Input
1
Output
1
Submitted Solution:
```
s = input()
s = s[::-1]
ans = 0
for i in range(len(s)):
if s[i] == '1':
ans += (2 ** i) ** 2
else:
ans *= 2
print(ans % (10**9+7))
``` | instruction | 0 | 99,738 | 14 | 199,476 |
No | output | 1 | 99,738 | 14 | 199,477 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As a tradition, every year before IOI all the members of Natalia Fan Club are invited to Malek Dance Club to have a fun night together. Malek Dance Club has 2n members and coincidentally Natalia Fan Club also has 2n members. Each member of MDC is assigned a unique id i from 0 to 2n - 1. The same holds for each member of NFC.
One of the parts of this tradition is one by one dance, where each member of MDC dances with a member of NFC. A dance pair is a pair of numbers (a, b) such that member a from MDC dances with member b from NFC.
The complexity of a pairs' assignment is the number of pairs of dancing pairs (a, b) and (c, d) such that a < c and b > d.
You are given a binary number of length n named x. We know that member i from MDC dances with member <image> from NFC. Your task is to calculate the complexity of this assignment modulo 1000000007 (109 + 7).
Expression <image> denotes applying «XOR» to numbers x and y. This operation exists in all modern programming languages, for example, in C++ and Java it denotes as «^», in Pascal — «xor».
Input
The first line of input contains a binary number x of lenght n, (1 ≤ n ≤ 100).
This number may contain leading zeros.
Output
Print the complexity of the given dance assignent modulo 1000000007 (109 + 7).
Examples
Input
11
Output
6
Input
01
Output
2
Input
1
Output
1
Submitted Solution:
```
s = input()
s = s[::-1]
ans = 0
for i in range(len(s)):
if s[i] == '1':
ans += 2 ** i
else:
ans *= 2
print(ans % (10**9+7))
``` | instruction | 0 | 99,739 | 14 | 199,478 |
No | output | 1 | 99,739 | 14 | 199,479 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kolya loves putting gnomes at the circle table and giving them coins, and Tanya loves studying triplets of gnomes, sitting in the vertexes of an equilateral triangle.
More formally, there are 3n gnomes sitting in a circle. Each gnome can have from 1 to 3 coins. Let's number the places in the order they occur in the circle by numbers from 0 to 3n - 1, let the gnome sitting on the i-th place have ai coins. If there is an integer i (0 ≤ i < n) such that ai + ai + n + ai + 2n ≠ 6, then Tanya is satisfied.
Count the number of ways to choose ai so that Tanya is satisfied. As there can be many ways of distributing coins, print the remainder of this number modulo 109 + 7. Two ways, a and b, are considered distinct if there is index i (0 ≤ i < 3n), such that ai ≠ bi (that is, some gnome got different number of coins in these two ways).
Input
A single line contains number n (1 ≤ n ≤ 105) — the number of the gnomes divided by three.
Output
Print a single number — the remainder of the number of variants of distributing coins that satisfy Tanya modulo 109 + 7.
Examples
Input
1
Output
20
Input
2
Output
680
Note
20 ways for n = 1 (gnome with index 0 sits on the top of the triangle, gnome 1 on the right vertex, gnome 2 on the left vertex): <image> | instruction | 0 | 99,829 | 14 | 199,658 |
Tags: combinatorics
Correct Solution:
```
n=int(input())
ans=27**n-7**n
mod=10**9+7
print(ans%mod)
``` | output | 1 | 99,829 | 14 | 199,659 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kolya loves putting gnomes at the circle table and giving them coins, and Tanya loves studying triplets of gnomes, sitting in the vertexes of an equilateral triangle.
More formally, there are 3n gnomes sitting in a circle. Each gnome can have from 1 to 3 coins. Let's number the places in the order they occur in the circle by numbers from 0 to 3n - 1, let the gnome sitting on the i-th place have ai coins. If there is an integer i (0 ≤ i < n) such that ai + ai + n + ai + 2n ≠ 6, then Tanya is satisfied.
Count the number of ways to choose ai so that Tanya is satisfied. As there can be many ways of distributing coins, print the remainder of this number modulo 109 + 7. Two ways, a and b, are considered distinct if there is index i (0 ≤ i < 3n), such that ai ≠ bi (that is, some gnome got different number of coins in these two ways).
Input
A single line contains number n (1 ≤ n ≤ 105) — the number of the gnomes divided by three.
Output
Print a single number — the remainder of the number of variants of distributing coins that satisfy Tanya modulo 109 + 7.
Examples
Input
1
Output
20
Input
2
Output
680
Note
20 ways for n = 1 (gnome with index 0 sits on the top of the triangle, gnome 1 on the right vertex, gnome 2 on the left vertex): <image> | instruction | 0 | 99,830 | 14 | 199,660 |
Tags: combinatorics
Correct Solution:
```
n=int(input())
p=9**n*3**n
print((p-7**n)%1000000007)
``` | output | 1 | 99,830 | 14 | 199,661 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kolya loves putting gnomes at the circle table and giving them coins, and Tanya loves studying triplets of gnomes, sitting in the vertexes of an equilateral triangle.
More formally, there are 3n gnomes sitting in a circle. Each gnome can have from 1 to 3 coins. Let's number the places in the order they occur in the circle by numbers from 0 to 3n - 1, let the gnome sitting on the i-th place have ai coins. If there is an integer i (0 ≤ i < n) such that ai + ai + n + ai + 2n ≠ 6, then Tanya is satisfied.
Count the number of ways to choose ai so that Tanya is satisfied. As there can be many ways of distributing coins, print the remainder of this number modulo 109 + 7. Two ways, a and b, are considered distinct if there is index i (0 ≤ i < 3n), such that ai ≠ bi (that is, some gnome got different number of coins in these two ways).
Input
A single line contains number n (1 ≤ n ≤ 105) — the number of the gnomes divided by three.
Output
Print a single number — the remainder of the number of variants of distributing coins that satisfy Tanya modulo 109 + 7.
Examples
Input
1
Output
20
Input
2
Output
680
Note
20 ways for n = 1 (gnome with index 0 sits on the top of the triangle, gnome 1 on the right vertex, gnome 2 on the left vertex): <image> | instruction | 0 | 99,831 | 14 | 199,662 |
Tags: combinatorics
Correct Solution:
```
n = input()
a, b = 20, 7
for i in range(1, int(n)):
a, b = a*27+b*20, b*7
a %= 1000000007
b %= 1000000007
print(a)
``` | output | 1 | 99,831 | 14 | 199,663 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kolya loves putting gnomes at the circle table and giving them coins, and Tanya loves studying triplets of gnomes, sitting in the vertexes of an equilateral triangle.
More formally, there are 3n gnomes sitting in a circle. Each gnome can have from 1 to 3 coins. Let's number the places in the order they occur in the circle by numbers from 0 to 3n - 1, let the gnome sitting on the i-th place have ai coins. If there is an integer i (0 ≤ i < n) such that ai + ai + n + ai + 2n ≠ 6, then Tanya is satisfied.
Count the number of ways to choose ai so that Tanya is satisfied. As there can be many ways of distributing coins, print the remainder of this number modulo 109 + 7. Two ways, a and b, are considered distinct if there is index i (0 ≤ i < 3n), such that ai ≠ bi (that is, some gnome got different number of coins in these two ways).
Input
A single line contains number n (1 ≤ n ≤ 105) — the number of the gnomes divided by three.
Output
Print a single number — the remainder of the number of variants of distributing coins that satisfy Tanya modulo 109 + 7.
Examples
Input
1
Output
20
Input
2
Output
680
Note
20 ways for n = 1 (gnome with index 0 sits on the top of the triangle, gnome 1 on the right vertex, gnome 2 on the left vertex): <image> | instruction | 0 | 99,832 | 14 | 199,664 |
Tags: combinatorics
Correct Solution:
```
'''
1 1 1
1 1 2
1 2 2
1 1 3
1 3 3
1 2 3
bad combo
3**(3*n)-badcombo
'''
n=int(input())
x=pow(3,3*n)-pow(7,n)
print(x%(10**9+7))
``` | output | 1 | 99,832 | 14 | 199,665 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kolya loves putting gnomes at the circle table and giving them coins, and Tanya loves studying triplets of gnomes, sitting in the vertexes of an equilateral triangle.
More formally, there are 3n gnomes sitting in a circle. Each gnome can have from 1 to 3 coins. Let's number the places in the order they occur in the circle by numbers from 0 to 3n - 1, let the gnome sitting on the i-th place have ai coins. If there is an integer i (0 ≤ i < n) such that ai + ai + n + ai + 2n ≠ 6, then Tanya is satisfied.
Count the number of ways to choose ai so that Tanya is satisfied. As there can be many ways of distributing coins, print the remainder of this number modulo 109 + 7. Two ways, a and b, are considered distinct if there is index i (0 ≤ i < 3n), such that ai ≠ bi (that is, some gnome got different number of coins in these two ways).
Input
A single line contains number n (1 ≤ n ≤ 105) — the number of the gnomes divided by three.
Output
Print a single number — the remainder of the number of variants of distributing coins that satisfy Tanya modulo 109 + 7.
Examples
Input
1
Output
20
Input
2
Output
680
Note
20 ways for n = 1 (gnome with index 0 sits on the top of the triangle, gnome 1 on the right vertex, gnome 2 on the left vertex): <image> | instruction | 0 | 99,833 | 14 | 199,666 |
Tags: combinatorics
Correct Solution:
```
n = int(input())
print((27 ** n - 7 ** n) % (10 ** 9 + 7))
``` | output | 1 | 99,833 | 14 | 199,667 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kolya loves putting gnomes at the circle table and giving them coins, and Tanya loves studying triplets of gnomes, sitting in the vertexes of an equilateral triangle.
More formally, there are 3n gnomes sitting in a circle. Each gnome can have from 1 to 3 coins. Let's number the places in the order they occur in the circle by numbers from 0 to 3n - 1, let the gnome sitting on the i-th place have ai coins. If there is an integer i (0 ≤ i < n) such that ai + ai + n + ai + 2n ≠ 6, then Tanya is satisfied.
Count the number of ways to choose ai so that Tanya is satisfied. As there can be many ways of distributing coins, print the remainder of this number modulo 109 + 7. Two ways, a and b, are considered distinct if there is index i (0 ≤ i < 3n), such that ai ≠ bi (that is, some gnome got different number of coins in these two ways).
Input
A single line contains number n (1 ≤ n ≤ 105) — the number of the gnomes divided by three.
Output
Print a single number — the remainder of the number of variants of distributing coins that satisfy Tanya modulo 109 + 7.
Examples
Input
1
Output
20
Input
2
Output
680
Note
20 ways for n = 1 (gnome with index 0 sits on the top of the triangle, gnome 1 on the right vertex, gnome 2 on the left vertex): <image> | instruction | 0 | 99,834 | 14 | 199,668 |
Tags: combinatorics
Correct Solution:
```
n = int(input())
mod = 10 ** 9 + 7
print((27 ** n - 7 ** n) % mod)
``` | output | 1 | 99,834 | 14 | 199,669 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kolya loves putting gnomes at the circle table and giving them coins, and Tanya loves studying triplets of gnomes, sitting in the vertexes of an equilateral triangle.
More formally, there are 3n gnomes sitting in a circle. Each gnome can have from 1 to 3 coins. Let's number the places in the order they occur in the circle by numbers from 0 to 3n - 1, let the gnome sitting on the i-th place have ai coins. If there is an integer i (0 ≤ i < n) such that ai + ai + n + ai + 2n ≠ 6, then Tanya is satisfied.
Count the number of ways to choose ai so that Tanya is satisfied. As there can be many ways of distributing coins, print the remainder of this number modulo 109 + 7. Two ways, a and b, are considered distinct if there is index i (0 ≤ i < 3n), such that ai ≠ bi (that is, some gnome got different number of coins in these two ways).
Input
A single line contains number n (1 ≤ n ≤ 105) — the number of the gnomes divided by three.
Output
Print a single number — the remainder of the number of variants of distributing coins that satisfy Tanya modulo 109 + 7.
Examples
Input
1
Output
20
Input
2
Output
680
Note
20 ways for n = 1 (gnome with index 0 sits on the top of the triangle, gnome 1 on the right vertex, gnome 2 on the left vertex): <image> | instruction | 0 | 99,835 | 14 | 199,670 |
Tags: combinatorics
Correct Solution:
```
n = int(input())
res = 1
res2 = 1
for i in range(3 * n):
res = (3 * res) % 1000000007
for i in range(n):
res2 = (7 * res2) % 1000000007
print((res - res2) % 1000000007)
``` | output | 1 | 99,835 | 14 | 199,671 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kolya loves putting gnomes at the circle table and giving them coins, and Tanya loves studying triplets of gnomes, sitting in the vertexes of an equilateral triangle.
More formally, there are 3n gnomes sitting in a circle. Each gnome can have from 1 to 3 coins. Let's number the places in the order they occur in the circle by numbers from 0 to 3n - 1, let the gnome sitting on the i-th place have ai coins. If there is an integer i (0 ≤ i < n) such that ai + ai + n + ai + 2n ≠ 6, then Tanya is satisfied.
Count the number of ways to choose ai so that Tanya is satisfied. As there can be many ways of distributing coins, print the remainder of this number modulo 109 + 7. Two ways, a and b, are considered distinct if there is index i (0 ≤ i < 3n), such that ai ≠ bi (that is, some gnome got different number of coins in these two ways).
Input
A single line contains number n (1 ≤ n ≤ 105) — the number of the gnomes divided by three.
Output
Print a single number — the remainder of the number of variants of distributing coins that satisfy Tanya modulo 109 + 7.
Examples
Input
1
Output
20
Input
2
Output
680
Note
20 ways for n = 1 (gnome with index 0 sits on the top of the triangle, gnome 1 on the right vertex, gnome 2 on the left vertex): <image> | instruction | 0 | 99,836 | 14 | 199,672 |
Tags: combinatorics
Correct Solution:
```
n = int(input())
ans = 27**n - 7**n
const = 10**9 + 7
print(ans%const)
``` | output | 1 | 99,836 | 14 | 199,673 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kolya loves putting gnomes at the circle table and giving them coins, and Tanya loves studying triplets of gnomes, sitting in the vertexes of an equilateral triangle.
More formally, there are 3n gnomes sitting in a circle. Each gnome can have from 1 to 3 coins. Let's number the places in the order they occur in the circle by numbers from 0 to 3n - 1, let the gnome sitting on the i-th place have ai coins. If there is an integer i (0 ≤ i < n) such that ai + ai + n + ai + 2n ≠ 6, then Tanya is satisfied.
Count the number of ways to choose ai so that Tanya is satisfied. As there can be many ways of distributing coins, print the remainder of this number modulo 109 + 7. Two ways, a and b, are considered distinct if there is index i (0 ≤ i < 3n), such that ai ≠ bi (that is, some gnome got different number of coins in these two ways).
Input
A single line contains number n (1 ≤ n ≤ 105) — the number of the gnomes divided by three.
Output
Print a single number — the remainder of the number of variants of distributing coins that satisfy Tanya modulo 109 + 7.
Examples
Input
1
Output
20
Input
2
Output
680
Note
20 ways for n = 1 (gnome with index 0 sits on the top of the triangle, gnome 1 on the right vertex, gnome 2 on the left vertex): <image>
Submitted Solution:
```
n = int(input())
r = 3**(n*3) - 7**n
print(r%1000000007)
``` | instruction | 0 | 99,837 | 14 | 199,674 |
Yes | output | 1 | 99,837 | 14 | 199,675 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kolya loves putting gnomes at the circle table and giving them coins, and Tanya loves studying triplets of gnomes, sitting in the vertexes of an equilateral triangle.
More formally, there are 3n gnomes sitting in a circle. Each gnome can have from 1 to 3 coins. Let's number the places in the order they occur in the circle by numbers from 0 to 3n - 1, let the gnome sitting on the i-th place have ai coins. If there is an integer i (0 ≤ i < n) such that ai + ai + n + ai + 2n ≠ 6, then Tanya is satisfied.
Count the number of ways to choose ai so that Tanya is satisfied. As there can be many ways of distributing coins, print the remainder of this number modulo 109 + 7. Two ways, a and b, are considered distinct if there is index i (0 ≤ i < 3n), such that ai ≠ bi (that is, some gnome got different number of coins in these two ways).
Input
A single line contains number n (1 ≤ n ≤ 105) — the number of the gnomes divided by three.
Output
Print a single number — the remainder of the number of variants of distributing coins that satisfy Tanya modulo 109 + 7.
Examples
Input
1
Output
20
Input
2
Output
680
Note
20 ways for n = 1 (gnome with index 0 sits on the top of the triangle, gnome 1 on the right vertex, gnome 2 on the left vertex): <image>
Submitted Solution:
```
inta = int(input());
intb = 3**(3*inta) - 7**(inta);
intb %= 1000000007;
print(intb);
``` | instruction | 0 | 99,838 | 14 | 199,676 |
Yes | output | 1 | 99,838 | 14 | 199,677 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kolya loves putting gnomes at the circle table and giving them coins, and Tanya loves studying triplets of gnomes, sitting in the vertexes of an equilateral triangle.
More formally, there are 3n gnomes sitting in a circle. Each gnome can have from 1 to 3 coins. Let's number the places in the order they occur in the circle by numbers from 0 to 3n - 1, let the gnome sitting on the i-th place have ai coins. If there is an integer i (0 ≤ i < n) such that ai + ai + n + ai + 2n ≠ 6, then Tanya is satisfied.
Count the number of ways to choose ai so that Tanya is satisfied. As there can be many ways of distributing coins, print the remainder of this number modulo 109 + 7. Two ways, a and b, are considered distinct if there is index i (0 ≤ i < 3n), such that ai ≠ bi (that is, some gnome got different number of coins in these two ways).
Input
A single line contains number n (1 ≤ n ≤ 105) — the number of the gnomes divided by three.
Output
Print a single number — the remainder of the number of variants of distributing coins that satisfy Tanya modulo 109 + 7.
Examples
Input
1
Output
20
Input
2
Output
680
Note
20 ways for n = 1 (gnome with index 0 sits on the top of the triangle, gnome 1 on the right vertex, gnome 2 on the left vertex): <image>
Submitted Solution:
```
mod_by = 10**9 + 7
n = int(input())
print((3**(3*n)-7**n) % mod_by)
``` | instruction | 0 | 99,839 | 14 | 199,678 |
Yes | output | 1 | 99,839 | 14 | 199,679 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kolya loves putting gnomes at the circle table and giving them coins, and Tanya loves studying triplets of gnomes, sitting in the vertexes of an equilateral triangle.
More formally, there are 3n gnomes sitting in a circle. Each gnome can have from 1 to 3 coins. Let's number the places in the order they occur in the circle by numbers from 0 to 3n - 1, let the gnome sitting on the i-th place have ai coins. If there is an integer i (0 ≤ i < n) such that ai + ai + n + ai + 2n ≠ 6, then Tanya is satisfied.
Count the number of ways to choose ai so that Tanya is satisfied. As there can be many ways of distributing coins, print the remainder of this number modulo 109 + 7. Two ways, a and b, are considered distinct if there is index i (0 ≤ i < 3n), such that ai ≠ bi (that is, some gnome got different number of coins in these two ways).
Input
A single line contains number n (1 ≤ n ≤ 105) — the number of the gnomes divided by three.
Output
Print a single number — the remainder of the number of variants of distributing coins that satisfy Tanya modulo 109 + 7.
Examples
Input
1
Output
20
Input
2
Output
680
Note
20 ways for n = 1 (gnome with index 0 sits on the top of the triangle, gnome 1 on the right vertex, gnome 2 on the left vertex): <image>
Submitted Solution:
```
n = int(input())
M = int(1e9 + 7)
print((pow(3, 3 * n, M) % M - pow(7, n, M) % M) % M)
``` | instruction | 0 | 99,840 | 14 | 199,680 |
Yes | output | 1 | 99,840 | 14 | 199,681 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kolya loves putting gnomes at the circle table and giving them coins, and Tanya loves studying triplets of gnomes, sitting in the vertexes of an equilateral triangle.
More formally, there are 3n gnomes sitting in a circle. Each gnome can have from 1 to 3 coins. Let's number the places in the order they occur in the circle by numbers from 0 to 3n - 1, let the gnome sitting on the i-th place have ai coins. If there is an integer i (0 ≤ i < n) such that ai + ai + n + ai + 2n ≠ 6, then Tanya is satisfied.
Count the number of ways to choose ai so that Tanya is satisfied. As there can be many ways of distributing coins, print the remainder of this number modulo 109 + 7. Two ways, a and b, are considered distinct if there is index i (0 ≤ i < 3n), such that ai ≠ bi (that is, some gnome got different number of coins in these two ways).
Input
A single line contains number n (1 ≤ n ≤ 105) — the number of the gnomes divided by three.
Output
Print a single number — the remainder of the number of variants of distributing coins that satisfy Tanya modulo 109 + 7.
Examples
Input
1
Output
20
Input
2
Output
680
Note
20 ways for n = 1 (gnome with index 0 sits on the top of the triangle, gnome 1 on the right vertex, gnome 2 on the left vertex): <image>
Submitted Solution:
```
n = int(input())
print(3**(3*n)-7**n % int(1e9+7))
``` | instruction | 0 | 99,841 | 14 | 199,682 |
No | output | 1 | 99,841 | 14 | 199,683 |
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