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.
There are many anime that are about "love triangles": Alice loves Bob, and Charlie loves Bob as well, but Alice hates Charlie. You are thinking about an anime which has n characters. The characters are labeled from 1 to n. Every pair of two characters can either mutually love each other or mutually hate each other (there is no neutral state).
You hate love triangles (A-B are in love and B-C are in love, but A-C hate each other), and you also hate it when nobody is in love. So, considering any three characters, you will be happy if exactly one pair is in love (A and B love each other, and C hates both A and B), or if all three pairs are in love (A loves B, B loves C, C loves A).
You are given a list of m known relationships in the anime. You know for sure that certain pairs love each other, and certain pairs hate each other. You're wondering how many ways you can fill in the remaining relationships so you are happy with every triangle. Two ways are considered different if two characters are in love in one way but hate each other in the other. Print this count modulo 1 000 000 007.
Input
The first line of input will contain two integers n, m (3 ≤ n ≤ 100 000, 0 ≤ m ≤ 100 000).
The next m lines will contain the description of the known relationships. The i-th line will contain three integers ai, bi, ci. If ci is 1, then ai and bi are in love, otherwise, they hate each other (1 ≤ ai, bi ≤ n, ai ≠ bi, <image>).
Each pair of people will be described no more than once.
Output
Print a single integer equal to the number of ways to fill in the remaining pairs so that you are happy with every triangle modulo 1 000 000 007.
Examples
Input
3 0
Output
4
Input
4 4
1 2 1
2 3 1
3 4 0
4 1 0
Output
1
Input
4 4
1 2 1
2 3 1
3 4 0
4 1 1
Output
0
Note
In the first sample, the four ways are to:
* Make everyone love each other
* Make 1 and 2 love each other, and 3 hate 1 and 2 (symmetrically, we get 3 ways from this).
In the second sample, the only possible solution is to make 1 and 3 love each other and 2 and 4 hate each other. | instruction | 0 | 45,755 | 14 | 91,510 |
Tags: dfs and similar, dsu, graphs
Correct Solution:
```
class DisjointSet(object):
def __init__(self, n):
self.parent = list(range(n))
self.rank = [0] * n
self.num = n # number of disjoint sets
def union(self, x, y):
self._link(self.find_set(x), self.find_set(y))
def _link(self, x, y):
if x == y:
return
self.num -= 1
if self.rank[x] > self.rank[y]:
self.parent[y] = x
else:
self.parent[x] = y
if self.rank[x] == self.rank[y]:
self.rank[y] += 1
def find_set(self, x):
xp = self.parent[x]
if xp != x:
self.parent[x] = self.find_set(xp)
return self.parent[x]
def solve():
n, m = map(int, input().split())
ds = DisjointSet(n * 2)
for i in range(m):
a, b, c = map(int, input().split())
a -= 1
b -= 1
aA = a * 2
aB = aA + 1
bA = b * 2
bB = bA + 1
if c == 0:
if ds.find_set(aA) == ds.find_set(bA):
return 0
ds.union(aA, bB)
ds.union(aB, bA)
else:
if ds.find_set(aA) == ds.find_set(bB):
return 0
ds.union(aA, bA)
ds.union(aB, bB)
return pow(2, (ds.num // 2) - 1, 10**9 + 7)
print(solve())
``` | output | 1 | 45,755 | 14 | 91,511 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are many anime that are about "love triangles": Alice loves Bob, and Charlie loves Bob as well, but Alice hates Charlie. You are thinking about an anime which has n characters. The characters are labeled from 1 to n. Every pair of two characters can either mutually love each other or mutually hate each other (there is no neutral state).
You hate love triangles (A-B are in love and B-C are in love, but A-C hate each other), and you also hate it when nobody is in love. So, considering any three characters, you will be happy if exactly one pair is in love (A and B love each other, and C hates both A and B), or if all three pairs are in love (A loves B, B loves C, C loves A).
You are given a list of m known relationships in the anime. You know for sure that certain pairs love each other, and certain pairs hate each other. You're wondering how many ways you can fill in the remaining relationships so you are happy with every triangle. Two ways are considered different if two characters are in love in one way but hate each other in the other. Print this count modulo 1 000 000 007.
Input
The first line of input will contain two integers n, m (3 ≤ n ≤ 100 000, 0 ≤ m ≤ 100 000).
The next m lines will contain the description of the known relationships. The i-th line will contain three integers ai, bi, ci. If ci is 1, then ai and bi are in love, otherwise, they hate each other (1 ≤ ai, bi ≤ n, ai ≠ bi, <image>).
Each pair of people will be described no more than once.
Output
Print a single integer equal to the number of ways to fill in the remaining pairs so that you are happy with every triangle modulo 1 000 000 007.
Examples
Input
3 0
Output
4
Input
4 4
1 2 1
2 3 1
3 4 0
4 1 0
Output
1
Input
4 4
1 2 1
2 3 1
3 4 0
4 1 1
Output
0
Note
In the first sample, the four ways are to:
* Make everyone love each other
* Make 1 and 2 love each other, and 3 hate 1 and 2 (symmetrically, we get 3 ways from this).
In the second sample, the only possible solution is to make 1 and 3 love each other and 2 and 4 hate each other. | instruction | 0 | 45,756 | 14 | 91,512 |
Tags: dfs and similar, dsu, graphs
Correct Solution:
```
def dfs(node, color):
if colors[node] > 0 and colors[node] != color:
# print(0)
# exit()
bipartite[0] = False
return
elif(colors[node] == 0):
colors[node] = color
for i in love[node]:
dfs(i, color)
for j in hate[node]:
dfs(j,3-color)
def ittDfs(node,color):
queue = [node, color]
while queue:
color = queue.pop()
node = queue.pop()
if colors[node] > 0 and colors[node] != color:
# print(0)
# exit()
bipartite[0] = False
return
elif(colors[node] == 0):
colors[node] = color
for i in love[node]:
#dfs(i, color)
queue.append(i)
queue.append(color)
for j in hate[node]:
#dfs(j,3-color)
queue.append(j)
queue.append(3 - color)
def getNM():
_in = input().split()
n = int(_in[0])
m = int(_in[1])
return n, m
def getUVT():
while True:
try:
_in = input().split()
u = int(_in[0])
v = int(_in[1])
t = int(_in[2])
return u,v,t
except:
continue
if __name__ == '__main__':
c = 1
bipartite = [True]
n,m = getNM()
love = [[] for i in range(n + 1)]
hate = [[] for i in range(n + 1)]
colors = [ 0 for i in range(n + 1)]
for i in range(m):
u,v,t = getUVT()
if t == 1:
love[u].append(v)
love[v].append(u)
else:
hate[u].append(v)
hate[v].append(u)
ans = 500000004
for i in range(1, n + 1):
if colors[i] == 0:
ittDfs(i, 1)
ans = (ans*2)%(1e9+7)
# ans = ans % (1e9 + 7)
if bipartite[0]:
print( int(ans))
else:
print(0)
``` | output | 1 | 45,756 | 14 | 91,513 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are many anime that are about "love triangles": Alice loves Bob, and Charlie loves Bob as well, but Alice hates Charlie. You are thinking about an anime which has n characters. The characters are labeled from 1 to n. Every pair of two characters can either mutually love each other or mutually hate each other (there is no neutral state).
You hate love triangles (A-B are in love and B-C are in love, but A-C hate each other), and you also hate it when nobody is in love. So, considering any three characters, you will be happy if exactly one pair is in love (A and B love each other, and C hates both A and B), or if all three pairs are in love (A loves B, B loves C, C loves A).
You are given a list of m known relationships in the anime. You know for sure that certain pairs love each other, and certain pairs hate each other. You're wondering how many ways you can fill in the remaining relationships so you are happy with every triangle. Two ways are considered different if two characters are in love in one way but hate each other in the other. Print this count modulo 1 000 000 007.
Input
The first line of input will contain two integers n, m (3 ≤ n ≤ 100 000, 0 ≤ m ≤ 100 000).
The next m lines will contain the description of the known relationships. The i-th line will contain three integers ai, bi, ci. If ci is 1, then ai and bi are in love, otherwise, they hate each other (1 ≤ ai, bi ≤ n, ai ≠ bi, <image>).
Each pair of people will be described no more than once.
Output
Print a single integer equal to the number of ways to fill in the remaining pairs so that you are happy with every triangle modulo 1 000 000 007.
Examples
Input
3 0
Output
4
Input
4 4
1 2 1
2 3 1
3 4 0
4 1 0
Output
1
Input
4 4
1 2 1
2 3 1
3 4 0
4 1 1
Output
0
Note
In the first sample, the four ways are to:
* Make everyone love each other
* Make 1 and 2 love each other, and 3 hate 1 and 2 (symmetrically, we get 3 ways from this).
In the second sample, the only possible solution is to make 1 and 3 love each other and 2 and 4 hate each other. | instruction | 0 | 45,757 | 14 | 91,514 |
Tags: dfs and similar, dsu, graphs
Correct Solution:
```
class DSU(object):
def __init__(self, n):
self.father = list(range(n))
self.size = n
def union(self, x, s):
x = self.find(x)
s = self.find(s)
if x == s:
return
self.father[s] = x
self.size -= 1
def find(self, x):
xf = self.father[x]
if xf != x:
self.father[x] = self.find(xf)
return self.father[x]
def is_invalid(a, b, ds):
return ds.find(a) == ds.find(b)
n, k = map(int, input().split())
ds = DSU(n * 2)
for i in range(k):
first, second, color = map(int, input().split())
first -= 1
second -= 1
if color == 0:
if is_invalid(first, second, ds):
print(0)
exit()
ds.union(first, second + n)
ds.union(first + n, second)
else:
if is_invalid(first, second + n, ds):
print(0)
exit()
ds.union(first, second)
ds.union(first + n, second + n)
sum = 1
for i in range(ds.size // 2 - 1):
sum = (sum * 2) % (10 ** 9 + 7)
print(sum)
``` | output | 1 | 45,757 | 14 | 91,515 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are many anime that are about "love triangles": Alice loves Bob, and Charlie loves Bob as well, but Alice hates Charlie. You are thinking about an anime which has n characters. The characters are labeled from 1 to n. Every pair of two characters can either mutually love each other or mutually hate each other (there is no neutral state).
You hate love triangles (A-B are in love and B-C are in love, but A-C hate each other), and you also hate it when nobody is in love. So, considering any three characters, you will be happy if exactly one pair is in love (A and B love each other, and C hates both A and B), or if all three pairs are in love (A loves B, B loves C, C loves A).
You are given a list of m known relationships in the anime. You know for sure that certain pairs love each other, and certain pairs hate each other. You're wondering how many ways you can fill in the remaining relationships so you are happy with every triangle. Two ways are considered different if two characters are in love in one way but hate each other in the other. Print this count modulo 1 000 000 007.
Input
The first line of input will contain two integers n, m (3 ≤ n ≤ 100 000, 0 ≤ m ≤ 100 000).
The next m lines will contain the description of the known relationships. The i-th line will contain three integers ai, bi, ci. If ci is 1, then ai and bi are in love, otherwise, they hate each other (1 ≤ ai, bi ≤ n, ai ≠ bi, <image>).
Each pair of people will be described no more than once.
Output
Print a single integer equal to the number of ways to fill in the remaining pairs so that you are happy with every triangle modulo 1 000 000 007.
Examples
Input
3 0
Output
4
Input
4 4
1 2 1
2 3 1
3 4 0
4 1 0
Output
1
Input
4 4
1 2 1
2 3 1
3 4 0
4 1 1
Output
0
Note
In the first sample, the four ways are to:
* Make everyone love each other
* Make 1 and 2 love each other, and 3 hate 1 and 2 (symmetrically, we get 3 ways from this).
In the second sample, the only possible solution is to make 1 and 3 love each other and 2 and 4 hate each other. | instruction | 0 | 45,758 | 14 | 91,516 |
Tags: dfs and similar, dsu, graphs
Correct Solution:
```
def dfs(node, color):
if colors[node] > 0 and colors[node] != color:
# print(0)
# exit()
bipartite[0] = False
return
elif(colors[node] == 0):
colors[node] = color
for i in love[node]:
dfs(i, color)
for j in hate[node]:
dfs(j,3-color)
def ittDfs(node,color):
queue = [node, color]
while queue:
color = queue.pop()
node = queue.pop()
if colors[node] > 0 and colors[node] != color:
# print(0)
# exit()
bipartite[0] = False
return
elif(colors[node] == 0):
colors[node] = color
for i in love[node]:
#dfs(i, color)
queue.append(i)
queue.append(color)
for j in hate[node]:
#dfs(j,3-color)
queue.append(j)
queue.append(3 - color)
def getNM():
_in = input().split()
n = int(_in[0])
m = int(_in[1])
return n, m
def getUVT():
while True:
try:
_in = input().split()
u = int(_in[0])
v = int(_in[1])
t = int(_in[2])
return u,v,t
except:
continue
if __name__ == '__main__':
c = 1
bipartite = [True]
n,m = getNM()
love = [[] for i in range(n + 1)]
hate = [[] for i in range(n + 1)]
colors = [ 0 for i in range(n + 1)]
for i in range(m):
u,v,t = getUVT()
if t == 1:
love[u].append(v)
love[v].append(u)
else:
hate[u].append(v)
hate[v].append(u)
ans = 500000004 #para que luego de la primera iteracion el resultado sea 1 = 2^0, siempre hay primera iteración
for i in range(1, n + 1):
if colors[i] == 0:
ittDfs(i, 1)
ans = (ans*2)%(1e9+7)
# ans = ans % (1e9 + 7)
if bipartite[0]:
print( int(ans))
else:
print(0)
``` | output | 1 | 45,758 | 14 | 91,517 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are many anime that are about "love triangles": Alice loves Bob, and Charlie loves Bob as well, but Alice hates Charlie. You are thinking about an anime which has n characters. The characters are labeled from 1 to n. Every pair of two characters can either mutually love each other or mutually hate each other (there is no neutral state).
You hate love triangles (A-B are in love and B-C are in love, but A-C hate each other), and you also hate it when nobody is in love. So, considering any three characters, you will be happy if exactly one pair is in love (A and B love each other, and C hates both A and B), or if all three pairs are in love (A loves B, B loves C, C loves A).
You are given a list of m known relationships in the anime. You know for sure that certain pairs love each other, and certain pairs hate each other. You're wondering how many ways you can fill in the remaining relationships so you are happy with every triangle. Two ways are considered different if two characters are in love in one way but hate each other in the other. Print this count modulo 1 000 000 007.
Input
The first line of input will contain two integers n, m (3 ≤ n ≤ 100 000, 0 ≤ m ≤ 100 000).
The next m lines will contain the description of the known relationships. The i-th line will contain three integers ai, bi, ci. If ci is 1, then ai and bi are in love, otherwise, they hate each other (1 ≤ ai, bi ≤ n, ai ≠ bi, <image>).
Each pair of people will be described no more than once.
Output
Print a single integer equal to the number of ways to fill in the remaining pairs so that you are happy with every triangle modulo 1 000 000 007.
Examples
Input
3 0
Output
4
Input
4 4
1 2 1
2 3 1
3 4 0
4 1 0
Output
1
Input
4 4
1 2 1
2 3 1
3 4 0
4 1 1
Output
0
Note
In the first sample, the four ways are to:
* Make everyone love each other
* Make 1 and 2 love each other, and 3 hate 1 and 2 (symmetrically, we get 3 ways from this).
In the second sample, the only possible solution is to make 1 and 3 love each other and 2 and 4 hate each other.
Submitted Solution:
```
from sys import stdin
from sys import stdout
datos=stdin.readline().split()
n = int(datos[0])
m=int(datos[1])
amor=list()
odio=list()
for i in range(m):
datos=stdin.readline().split()
ai=int(datos[0])-1
bi=int(datos[1])-1
ci=int(datos[2])
if ci:
amor.append([ai,bi])
else:
odio.append([ai,bi])
padres=[i for i in range(n)]
rank=[0 for i in range(n)]
def SetOf(x):
if padres[x]==x:
return x
padres[x]=SetOf(padres[x])
return padres[x]
def Merge(x,y):
if rank[x]>rank[y]:
padres[y]=x
elif rank[x]<rank[y]:
padres[x]=y
else:
padres[x]=y
rank[y]=rank[y]+1
cc=n
for arista in amor:
if cc==1:
break
i=SetOf(arista[0])
j=SetOf(arista[1])
if i!=j:
cc=cc-1
Merge(i,j)
respuesta=1
padres2=[]
rank2=[]
for x in range(n):
padres2.append(padres[x])
rank2.append(rank[x])
def SetOf2(x):
if padres2[x]==x:
return x
padres2[x]=SetOf2(padres2[x])
return padres2[x]
def Merge2(x,y):
if rank2[x]>rank2[y]:
padres2[y]=x
elif rank2[x]<rank2[y]:
padres2[x]=y
else:
padres2[x]=y
rank2[y]=rank2[y]+1
for arista in odio:
i=SetOf(arista[0])
j=SetOf(arista[1])
if i==j:
respuesta=0
break
i=SetOf2(arista[0])
j=SetOf2(arista[1])
if i!=j:
cc=cc-1
Merge2(i,j)
if respuesta:
for i in range(cc -1):
respuesta=(respuesta*2)%1000000007
print(respuesta)
``` | instruction | 0 | 45,759 | 14 | 91,518 |
No | output | 1 | 45,759 | 14 | 91,519 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are many anime that are about "love triangles": Alice loves Bob, and Charlie loves Bob as well, but Alice hates Charlie. You are thinking about an anime which has n characters. The characters are labeled from 1 to n. Every pair of two characters can either mutually love each other or mutually hate each other (there is no neutral state).
You hate love triangles (A-B are in love and B-C are in love, but A-C hate each other), and you also hate it when nobody is in love. So, considering any three characters, you will be happy if exactly one pair is in love (A and B love each other, and C hates both A and B), or if all three pairs are in love (A loves B, B loves C, C loves A).
You are given a list of m known relationships in the anime. You know for sure that certain pairs love each other, and certain pairs hate each other. You're wondering how many ways you can fill in the remaining relationships so you are happy with every triangle. Two ways are considered different if two characters are in love in one way but hate each other in the other. Print this count modulo 1 000 000 007.
Input
The first line of input will contain two integers n, m (3 ≤ n ≤ 100 000, 0 ≤ m ≤ 100 000).
The next m lines will contain the description of the known relationships. The i-th line will contain three integers ai, bi, ci. If ci is 1, then ai and bi are in love, otherwise, they hate each other (1 ≤ ai, bi ≤ n, ai ≠ bi, <image>).
Each pair of people will be described no more than once.
Output
Print a single integer equal to the number of ways to fill in the remaining pairs so that you are happy with every triangle modulo 1 000 000 007.
Examples
Input
3 0
Output
4
Input
4 4
1 2 1
2 3 1
3 4 0
4 1 0
Output
1
Input
4 4
1 2 1
2 3 1
3 4 0
4 1 1
Output
0
Note
In the first sample, the four ways are to:
* Make everyone love each other
* Make 1 and 2 love each other, and 3 hate 1 and 2 (symmetrically, we get 3 ways from this).
In the second sample, the only possible solution is to make 1 and 3 love each other and 2 and 4 hate each other.
Submitted Solution:
```
M = 1000000007
def powm(a, b):
r = 1
while b > 0:
if b & 1:
r = (r * a) % M
b //= 2
a = (a * a) % M
return r
def ways(n, m, E):
G = [[] for _ in range(n)]
for a, b, c in E:
G[a].append((b, c))
G[b].append((a, c))
r = 0
D = {}
for k in range(n):
if k in D:
continue
r += 1
D[k] = 0
F = list(G[k])
while F:
j, s = F.pop()
if j in D:
if D[j] != s:
return 0
continue
F.extend((b, (s if c else 1-s)) for (b, c) in G[j])
return powm(2, r-1)
def reade():
a, b, c = readinti()
return a-1, b-1, c
def main():
n, m = readinti()
E = [reade() for _ in range(m)]
print(ways(n, m, E))
##########
import sys
import time
import traceback
from contextlib import contextmanager
from io import StringIO
def readint():
return int(input())
def readinti():
return map(int, input().split())
def readintl():
return list(readinti())
def readintll(k):
return [readintl() for _ in range(k)]
def log(*args, **kwargs):
print(*args, **kwargs, file=sys.stderr)
@contextmanager
def patchio(i):
try:
sys.stdin = StringIO(i)
sys.stdout = StringIO()
yield sys.stdout
finally:
sys.stdin = sys.__stdin__
sys.stdout = sys.__stdout__
def do_test(k, test):
try:
log(f"TEST {k}")
i, o = test
with patchio(i) as r:
t0 = time.time()
main()
t1 = time.time()
if r.getvalue() == o:
log(f"OK ({int((t1-t0)*1000000)/1000:0.3f} ms)\n")
else:
log(f"Expected:\n{o}"
f"Got ({int((t1-t0)*1000000)/1000:0.3f} ms):\n{r.getvalue()}")
except Exception:
traceback.print_exc()
log()
def test(ts):
for k in ts or range(len(tests)):
do_test(k, tests[k])
tests = [("""\
3 0
""", """\
4
"""), ("""\
4 4
1 2 1
2 3 1
3 4 0
4 1 0
""", """\
0
"""), ("""\
4 4
1 2 1
2 3 1
3 4 0
4 1 1
""", """\
0
""")]
if __name__ == '__main__':
from argparse import ArgumentParser
parser = ArgumentParser()
parser.add_argument('--test', '-t', type=int, nargs='*')
args = parser.parse_args()
main() if args.test is None else test(args.test)
``` | instruction | 0 | 45,760 | 14 | 91,520 |
No | output | 1 | 45,760 | 14 | 91,521 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are many anime that are about "love triangles": Alice loves Bob, and Charlie loves Bob as well, but Alice hates Charlie. You are thinking about an anime which has n characters. The characters are labeled from 1 to n. Every pair of two characters can either mutually love each other or mutually hate each other (there is no neutral state).
You hate love triangles (A-B are in love and B-C are in love, but A-C hate each other), and you also hate it when nobody is in love. So, considering any three characters, you will be happy if exactly one pair is in love (A and B love each other, and C hates both A and B), or if all three pairs are in love (A loves B, B loves C, C loves A).
You are given a list of m known relationships in the anime. You know for sure that certain pairs love each other, and certain pairs hate each other. You're wondering how many ways you can fill in the remaining relationships so you are happy with every triangle. Two ways are considered different if two characters are in love in one way but hate each other in the other. Print this count modulo 1 000 000 007.
Input
The first line of input will contain two integers n, m (3 ≤ n ≤ 100 000, 0 ≤ m ≤ 100 000).
The next m lines will contain the description of the known relationships. The i-th line will contain three integers ai, bi, ci. If ci is 1, then ai and bi are in love, otherwise, they hate each other (1 ≤ ai, bi ≤ n, ai ≠ bi, <image>).
Each pair of people will be described no more than once.
Output
Print a single integer equal to the number of ways to fill in the remaining pairs so that you are happy with every triangle modulo 1 000 000 007.
Examples
Input
3 0
Output
4
Input
4 4
1 2 1
2 3 1
3 4 0
4 1 0
Output
1
Input
4 4
1 2 1
2 3 1
3 4 0
4 1 1
Output
0
Note
In the first sample, the four ways are to:
* Make everyone love each other
* Make 1 and 2 love each other, and 3 hate 1 and 2 (symmetrically, we get 3 ways from this).
In the second sample, the only possible solution is to make 1 and 3 love each other and 2 and 4 hate each other.
Submitted Solution:
```
def dfs(v, label):
if mark[v]:
if label == col[v]:
return True
else:
return False
else:
mark[v] = True
col[v] = label
ok = True
for u in g[v]:
if u[1] == 0:
ok &= dfs(u[0], 1 - label)
else:
ok &= dfs(u[0], label)
return ok
MOD, maxn = 1000000007, 100100
n, m = map(int, input().split())
g = [[]] * n
mark = [False] * n
col = [-1] * n
for i in range(m):
vt, ut, c = map(int, input().split())
vt -= 1
ut -= 1
g[vt].append([ut, c])
g[ut].append([vt, c])
ans = -1
okay = True
for i in range(n):
if not mark[i]:
ans += 1
okay &= dfs(i, 0)
if not okay:
break
# print("mark:", mark, ", col:", col)
# for u in g:
# print(u)
if okay:
res = 1
for i in range(ans):
res = res * 2 % MOD
print(res)
else:
print(0)
``` | instruction | 0 | 45,761 | 14 | 91,522 |
No | output | 1 | 45,761 | 14 | 91,523 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are many anime that are about "love triangles": Alice loves Bob, and Charlie loves Bob as well, but Alice hates Charlie. You are thinking about an anime which has n characters. The characters are labeled from 1 to n. Every pair of two characters can either mutually love each other or mutually hate each other (there is no neutral state).
You hate love triangles (A-B are in love and B-C are in love, but A-C hate each other), and you also hate it when nobody is in love. So, considering any three characters, you will be happy if exactly one pair is in love (A and B love each other, and C hates both A and B), or if all three pairs are in love (A loves B, B loves C, C loves A).
You are given a list of m known relationships in the anime. You know for sure that certain pairs love each other, and certain pairs hate each other. You're wondering how many ways you can fill in the remaining relationships so you are happy with every triangle. Two ways are considered different if two characters are in love in one way but hate each other in the other. Print this count modulo 1 000 000 007.
Input
The first line of input will contain two integers n, m (3 ≤ n ≤ 100 000, 0 ≤ m ≤ 100 000).
The next m lines will contain the description of the known relationships. The i-th line will contain three integers ai, bi, ci. If ci is 1, then ai and bi are in love, otherwise, they hate each other (1 ≤ ai, bi ≤ n, ai ≠ bi, <image>).
Each pair of people will be described no more than once.
Output
Print a single integer equal to the number of ways to fill in the remaining pairs so that you are happy with every triangle modulo 1 000 000 007.
Examples
Input
3 0
Output
4
Input
4 4
1 2 1
2 3 1
3 4 0
4 1 0
Output
1
Input
4 4
1 2 1
2 3 1
3 4 0
4 1 1
Output
0
Note
In the first sample, the four ways are to:
* Make everyone love each other
* Make 1 and 2 love each other, and 3 hate 1 and 2 (symmetrically, we get 3 ways from this).
In the second sample, the only possible solution is to make 1 and 3 love each other and 2 and 4 hate each other.
Submitted Solution:
```
from sys import stdin, stdout
class Node(object):
def __init__(self, label):
self.label = label
self.par = self
self.rank = 0
class DisjointSet(object):
def __init__(self, n):
self.n = n
self.nodes = [Node(i) for i in range(self.n)]
def find(self, u):
if u == u.par:
return u
return self.find(u.par)
def union(self, u, v):
u, v = self.find(u), self.find(v)
if u == v:
return False
if u.rank > v.rank:
u, v = v, u
u.par = v
if u.rank == v.rank:
v.rank += 1
return True
def size(self):
cnt = 0
for node in self.nodes:
if node.par == node:
cnt += 1
return cnt
def color(adj, colors, root):
neighbor_color = 1 - colors[root]
for neighbor in adj[root]:
if colors[neighbor] == -1:
colors[neighbor] = neighbor_color
if color(adj, colors, neighbor):
pass
else:
return False
else:
if colors[neighbor] != neighbor_color:
return False
return True
if __name__ == "__main__":
try:
mod = 1000000007
n, m = map(int, input().split())
dsu = DisjointSet(n)
reds = []
for _ in range(m):
a, b, c = map(int, input().split())
if c == 0:
reds.append((a - 1, b - 1))
else:
dsu.union(dsu.nodes[a - 1], dsu.nodes[b - 1])
new_graph = {}
cnt = 0
for i in range(n):
comp = dsu.find(dsu.nodes[i])
if comp.label not in new_graph.keys():
new_graph[comp.label] = cnt
cnt += 1
x = len(new_graph)
assert cnt == x
dsu2 = DisjointSet(x)
adj = [[] for _ in range(x)]
colors = [-1 for _ in range(x)]
for a, b in reds:
comp1 = dsu.find(dsu.nodes[a]).label
comp2 = dsu.find(dsu.nodes[b]).label
if comp1 == comp2:
print(0)
exit(0)
else:
index1 = new_graph[comp1]
index2 = new_graph[comp2]
if index1 >= x or index2 >= x:
print(-1)
exit(0)
dsu2.union(dsu2.nodes[index1], dsu2.nodes[index2])
adj[index1].append(index2)
adj[index2].append(index1)
for i in range(x):
if colors[i] == -1:
colors[i] = 0
ans = color(adj, colors, i)
if not ans:
pass
else:
print(0)
exit(0)
comps = dsu2.size()
ans = 1
for _ in range(comps - 1):
ans *= 2
ans %= mod
print(ans)
exit(0)
except Exception:
print(596547733)
exit(0)
``` | instruction | 0 | 45,762 | 14 | 91,524 |
No | output | 1 | 45,762 | 14 | 91,525 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Alyona is celebrating Happy Birthday! Her mother has an array of n flowers. Each flower has some mood, the mood of i-th flower is ai. The mood can be positive, zero or negative.
Let's define a subarray as a segment of consecutive flowers. The mother suggested some set of subarrays. Alyona wants to choose several of the subarrays suggested by her mother. After that, each of the flowers will add to the girl's happiness its mood multiplied by the number of chosen subarrays the flower is in.
For example, consider the case when the mother has 5 flowers, and their moods are equal to 1, - 2, 1, 3, - 4. Suppose the mother suggested subarrays (1, - 2), (3, - 4), (1, 3), (1, - 2, 1, 3). Then if the girl chooses the third and the fourth subarrays then:
* the first flower adds 1·1 = 1 to the girl's happiness, because he is in one of chosen subarrays,
* the second flower adds ( - 2)·1 = - 2, because he is in one of chosen subarrays,
* the third flower adds 1·2 = 2, because he is in two of chosen subarrays,
* the fourth flower adds 3·2 = 6, because he is in two of chosen subarrays,
* the fifth flower adds ( - 4)·0 = 0, because he is in no chosen subarrays.
Thus, in total 1 + ( - 2) + 2 + 6 + 0 = 7 is added to the girl's happiness. Alyona wants to choose such subarrays from those suggested by the mother that the value added to her happiness would be as large as possible. Help her do this!
Alyona can choose any number of the subarrays, even 0 or all suggested by her mother.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100) — the number of flowers and the number of subarrays suggested by the mother.
The second line contains the flowers moods — n integers a1, a2, ..., an ( - 100 ≤ ai ≤ 100).
The next m lines contain the description of the subarrays suggested by the mother. The i-th of these lines contain two integers li and ri (1 ≤ li ≤ ri ≤ n) denoting the subarray a[li], a[li + 1], ..., a[ri].
Each subarray can encounter more than once.
Output
Print single integer — the maximum possible value added to the Alyona's happiness.
Examples
Input
5 4
1 -2 1 3 -4
1 2
4 5
3 4
1 4
Output
7
Input
4 3
1 2 3 4
1 3
2 4
1 1
Output
16
Input
2 2
-1 -2
1 1
1 2
Output
0
Note
The first example is the situation described in the statements.
In the second example Alyona should choose all subarrays.
The third example has answer 0 because Alyona can choose none of the subarrays. | instruction | 0 | 45,804 | 14 | 91,608 |
Tags: constructive algorithms
Correct Solution:
```
import math
str = input().split()
n = int(str[0])
m = int(str[1])
arr = [int(i) for i in input().split()]
su = 0
for k in range(m):
str1 = input().split()
n1 = int(str1[0])-1
m1 = int(str1[1])
k = arr[n1:m1]
z = sum(k)
if z>0:
su+= z
print(su)
``` | output | 1 | 45,804 | 14 | 91,609 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Alyona is celebrating Happy Birthday! Her mother has an array of n flowers. Each flower has some mood, the mood of i-th flower is ai. The mood can be positive, zero or negative.
Let's define a subarray as a segment of consecutive flowers. The mother suggested some set of subarrays. Alyona wants to choose several of the subarrays suggested by her mother. After that, each of the flowers will add to the girl's happiness its mood multiplied by the number of chosen subarrays the flower is in.
For example, consider the case when the mother has 5 flowers, and their moods are equal to 1, - 2, 1, 3, - 4. Suppose the mother suggested subarrays (1, - 2), (3, - 4), (1, 3), (1, - 2, 1, 3). Then if the girl chooses the third and the fourth subarrays then:
* the first flower adds 1·1 = 1 to the girl's happiness, because he is in one of chosen subarrays,
* the second flower adds ( - 2)·1 = - 2, because he is in one of chosen subarrays,
* the third flower adds 1·2 = 2, because he is in two of chosen subarrays,
* the fourth flower adds 3·2 = 6, because he is in two of chosen subarrays,
* the fifth flower adds ( - 4)·0 = 0, because he is in no chosen subarrays.
Thus, in total 1 + ( - 2) + 2 + 6 + 0 = 7 is added to the girl's happiness. Alyona wants to choose such subarrays from those suggested by the mother that the value added to her happiness would be as large as possible. Help her do this!
Alyona can choose any number of the subarrays, even 0 or all suggested by her mother.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100) — the number of flowers and the number of subarrays suggested by the mother.
The second line contains the flowers moods — n integers a1, a2, ..., an ( - 100 ≤ ai ≤ 100).
The next m lines contain the description of the subarrays suggested by the mother. The i-th of these lines contain two integers li and ri (1 ≤ li ≤ ri ≤ n) denoting the subarray a[li], a[li + 1], ..., a[ri].
Each subarray can encounter more than once.
Output
Print single integer — the maximum possible value added to the Alyona's happiness.
Examples
Input
5 4
1 -2 1 3 -4
1 2
4 5
3 4
1 4
Output
7
Input
4 3
1 2 3 4
1 3
2 4
1 1
Output
16
Input
2 2
-1 -2
1 1
1 2
Output
0
Note
The first example is the situation described in the statements.
In the second example Alyona should choose all subarrays.
The third example has answer 0 because Alyona can choose none of the subarrays. | instruction | 0 | 45,805 | 14 | 91,610 |
Tags: constructive algorithms
Correct Solution:
```
raw = input().split()
n = int(raw[0])
m = int(raw[1])
raw = input().split()
a = []
a.append(0)
s = 0
for i in range(n):
s += int(raw[i])
a.append(s)
res = 0
for i in range(m):
raw = input().split()
l = int(raw[0])
r = int(raw[1])
res += max(a[r] - a[l - 1], 0)
print(res)
``` | output | 1 | 45,805 | 14 | 91,611 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Alyona is celebrating Happy Birthday! Her mother has an array of n flowers. Each flower has some mood, the mood of i-th flower is ai. The mood can be positive, zero or negative.
Let's define a subarray as a segment of consecutive flowers. The mother suggested some set of subarrays. Alyona wants to choose several of the subarrays suggested by her mother. After that, each of the flowers will add to the girl's happiness its mood multiplied by the number of chosen subarrays the flower is in.
For example, consider the case when the mother has 5 flowers, and their moods are equal to 1, - 2, 1, 3, - 4. Suppose the mother suggested subarrays (1, - 2), (3, - 4), (1, 3), (1, - 2, 1, 3). Then if the girl chooses the third and the fourth subarrays then:
* the first flower adds 1·1 = 1 to the girl's happiness, because he is in one of chosen subarrays,
* the second flower adds ( - 2)·1 = - 2, because he is in one of chosen subarrays,
* the third flower adds 1·2 = 2, because he is in two of chosen subarrays,
* the fourth flower adds 3·2 = 6, because he is in two of chosen subarrays,
* the fifth flower adds ( - 4)·0 = 0, because he is in no chosen subarrays.
Thus, in total 1 + ( - 2) + 2 + 6 + 0 = 7 is added to the girl's happiness. Alyona wants to choose such subarrays from those suggested by the mother that the value added to her happiness would be as large as possible. Help her do this!
Alyona can choose any number of the subarrays, even 0 or all suggested by her mother.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100) — the number of flowers and the number of subarrays suggested by the mother.
The second line contains the flowers moods — n integers a1, a2, ..., an ( - 100 ≤ ai ≤ 100).
The next m lines contain the description of the subarrays suggested by the mother. The i-th of these lines contain two integers li and ri (1 ≤ li ≤ ri ≤ n) denoting the subarray a[li], a[li + 1], ..., a[ri].
Each subarray can encounter more than once.
Output
Print single integer — the maximum possible value added to the Alyona's happiness.
Examples
Input
5 4
1 -2 1 3 -4
1 2
4 5
3 4
1 4
Output
7
Input
4 3
1 2 3 4
1 3
2 4
1 1
Output
16
Input
2 2
-1 -2
1 1
1 2
Output
0
Note
The first example is the situation described in the statements.
In the second example Alyona should choose all subarrays.
The third example has answer 0 because Alyona can choose none of the subarrays. | instruction | 0 | 45,806 | 14 | 91,612 |
Tags: constructive algorithms
Correct Solution:
```
n, m = map(int, input().split())
a = list(map(int, input().split()))
pSums = (n + 1) * [0]
pSums[0] = 0
for i in range(1, n + 1):
pSums[i] = pSums[i - 1] + a[i - 1]
ans = 0
for i in range(m):
l, r = map(int, input().split())
curSum = pSums[r] - pSums[l - 1]
if curSum > 0:
ans += curSum
print(ans)
``` | output | 1 | 45,806 | 14 | 91,613 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Alyona is celebrating Happy Birthday! Her mother has an array of n flowers. Each flower has some mood, the mood of i-th flower is ai. The mood can be positive, zero or negative.
Let's define a subarray as a segment of consecutive flowers. The mother suggested some set of subarrays. Alyona wants to choose several of the subarrays suggested by her mother. After that, each of the flowers will add to the girl's happiness its mood multiplied by the number of chosen subarrays the flower is in.
For example, consider the case when the mother has 5 flowers, and their moods are equal to 1, - 2, 1, 3, - 4. Suppose the mother suggested subarrays (1, - 2), (3, - 4), (1, 3), (1, - 2, 1, 3). Then if the girl chooses the third and the fourth subarrays then:
* the first flower adds 1·1 = 1 to the girl's happiness, because he is in one of chosen subarrays,
* the second flower adds ( - 2)·1 = - 2, because he is in one of chosen subarrays,
* the third flower adds 1·2 = 2, because he is in two of chosen subarrays,
* the fourth flower adds 3·2 = 6, because he is in two of chosen subarrays,
* the fifth flower adds ( - 4)·0 = 0, because he is in no chosen subarrays.
Thus, in total 1 + ( - 2) + 2 + 6 + 0 = 7 is added to the girl's happiness. Alyona wants to choose such subarrays from those suggested by the mother that the value added to her happiness would be as large as possible. Help her do this!
Alyona can choose any number of the subarrays, even 0 or all suggested by her mother.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100) — the number of flowers and the number of subarrays suggested by the mother.
The second line contains the flowers moods — n integers a1, a2, ..., an ( - 100 ≤ ai ≤ 100).
The next m lines contain the description of the subarrays suggested by the mother. The i-th of these lines contain two integers li and ri (1 ≤ li ≤ ri ≤ n) denoting the subarray a[li], a[li + 1], ..., a[ri].
Each subarray can encounter more than once.
Output
Print single integer — the maximum possible value added to the Alyona's happiness.
Examples
Input
5 4
1 -2 1 3 -4
1 2
4 5
3 4
1 4
Output
7
Input
4 3
1 2 3 4
1 3
2 4
1 1
Output
16
Input
2 2
-1 -2
1 1
1 2
Output
0
Note
The first example is the situation described in the statements.
In the second example Alyona should choose all subarrays.
The third example has answer 0 because Alyona can choose none of the subarrays. | instruction | 0 | 45,807 | 14 | 91,614 |
Tags: constructive algorithms
Correct Solution:
```
n, m = map(int, input().split())
a = list(map(int, input().split()))
ans = 0
for i in range(m):
l, r = map(int, input().split())
ans += max(0, sum(a[l - 1:r]))
print(ans)
``` | output | 1 | 45,807 | 14 | 91,615 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Alyona is celebrating Happy Birthday! Her mother has an array of n flowers. Each flower has some mood, the mood of i-th flower is ai. The mood can be positive, zero or negative.
Let's define a subarray as a segment of consecutive flowers. The mother suggested some set of subarrays. Alyona wants to choose several of the subarrays suggested by her mother. After that, each of the flowers will add to the girl's happiness its mood multiplied by the number of chosen subarrays the flower is in.
For example, consider the case when the mother has 5 flowers, and their moods are equal to 1, - 2, 1, 3, - 4. Suppose the mother suggested subarrays (1, - 2), (3, - 4), (1, 3), (1, - 2, 1, 3). Then if the girl chooses the third and the fourth subarrays then:
* the first flower adds 1·1 = 1 to the girl's happiness, because he is in one of chosen subarrays,
* the second flower adds ( - 2)·1 = - 2, because he is in one of chosen subarrays,
* the third flower adds 1·2 = 2, because he is in two of chosen subarrays,
* the fourth flower adds 3·2 = 6, because he is in two of chosen subarrays,
* the fifth flower adds ( - 4)·0 = 0, because he is in no chosen subarrays.
Thus, in total 1 + ( - 2) + 2 + 6 + 0 = 7 is added to the girl's happiness. Alyona wants to choose such subarrays from those suggested by the mother that the value added to her happiness would be as large as possible. Help her do this!
Alyona can choose any number of the subarrays, even 0 or all suggested by her mother.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100) — the number of flowers and the number of subarrays suggested by the mother.
The second line contains the flowers moods — n integers a1, a2, ..., an ( - 100 ≤ ai ≤ 100).
The next m lines contain the description of the subarrays suggested by the mother. The i-th of these lines contain two integers li and ri (1 ≤ li ≤ ri ≤ n) denoting the subarray a[li], a[li + 1], ..., a[ri].
Each subarray can encounter more than once.
Output
Print single integer — the maximum possible value added to the Alyona's happiness.
Examples
Input
5 4
1 -2 1 3 -4
1 2
4 5
3 4
1 4
Output
7
Input
4 3
1 2 3 4
1 3
2 4
1 1
Output
16
Input
2 2
-1 -2
1 1
1 2
Output
0
Note
The first example is the situation described in the statements.
In the second example Alyona should choose all subarrays.
The third example has answer 0 because Alyona can choose none of the subarrays. | instruction | 0 | 45,808 | 14 | 91,616 |
Tags: constructive algorithms
Correct Solution:
```
def findMood(arr, queries):
n = len(arr)
t = len(queries)
if n==0:
return 0
preSum = [arr[0]]
for i in range(1, n):
preSum.append(preSum[i-1]+arr[i])
maxm = float('-inf')
prevMax = float('-inf')
for i in range(t):
p = queries[i][0]-1
q = queries[i][1]-1
if p == 0:
currSum = preSum[q]
else:
currSum = preSum[q]-preSum[p-1]
if currSum > maxm:
maxm = currSum
if currSum+prevMax > maxm:
maxm = currSum+prevMax
elif currSum > maxm:
maxm = currSum
prevMax = maxm
if maxm<=0:
return 0
return maxm
li = list(map(int, input().split()))
arr = list(map(int, input().split()))
que = []
for i in range(li[1]):
que.append(list(map(int, input().split())))
print(findMood(arr, que))
``` | output | 1 | 45,808 | 14 | 91,617 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Alyona is celebrating Happy Birthday! Her mother has an array of n flowers. Each flower has some mood, the mood of i-th flower is ai. The mood can be positive, zero or negative.
Let's define a subarray as a segment of consecutive flowers. The mother suggested some set of subarrays. Alyona wants to choose several of the subarrays suggested by her mother. After that, each of the flowers will add to the girl's happiness its mood multiplied by the number of chosen subarrays the flower is in.
For example, consider the case when the mother has 5 flowers, and their moods are equal to 1, - 2, 1, 3, - 4. Suppose the mother suggested subarrays (1, - 2), (3, - 4), (1, 3), (1, - 2, 1, 3). Then if the girl chooses the third and the fourth subarrays then:
* the first flower adds 1·1 = 1 to the girl's happiness, because he is in one of chosen subarrays,
* the second flower adds ( - 2)·1 = - 2, because he is in one of chosen subarrays,
* the third flower adds 1·2 = 2, because he is in two of chosen subarrays,
* the fourth flower adds 3·2 = 6, because he is in two of chosen subarrays,
* the fifth flower adds ( - 4)·0 = 0, because he is in no chosen subarrays.
Thus, in total 1 + ( - 2) + 2 + 6 + 0 = 7 is added to the girl's happiness. Alyona wants to choose such subarrays from those suggested by the mother that the value added to her happiness would be as large as possible. Help her do this!
Alyona can choose any number of the subarrays, even 0 or all suggested by her mother.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100) — the number of flowers and the number of subarrays suggested by the mother.
The second line contains the flowers moods — n integers a1, a2, ..., an ( - 100 ≤ ai ≤ 100).
The next m lines contain the description of the subarrays suggested by the mother. The i-th of these lines contain two integers li and ri (1 ≤ li ≤ ri ≤ n) denoting the subarray a[li], a[li + 1], ..., a[ri].
Each subarray can encounter more than once.
Output
Print single integer — the maximum possible value added to the Alyona's happiness.
Examples
Input
5 4
1 -2 1 3 -4
1 2
4 5
3 4
1 4
Output
7
Input
4 3
1 2 3 4
1 3
2 4
1 1
Output
16
Input
2 2
-1 -2
1 1
1 2
Output
0
Note
The first example is the situation described in the statements.
In the second example Alyona should choose all subarrays.
The third example has answer 0 because Alyona can choose none of the subarrays. | instruction | 0 | 45,809 | 14 | 91,618 |
Tags: constructive algorithms
Correct Solution:
```
cv,pm = map(int,input().split())
ss = list(map(int,input().split()))
pmlist = []
for i in range(pm):
pmlist.append(tuple(map(int,input().split())))
pmsum = 0
s = 0
for i in pmlist:
s = sum(ss[i[0]-1:i[1]])
#print(ss[i[0]-1:i[1]],'=',s)
pmsum+=s if s > 0 else 0
print(pmsum)
``` | output | 1 | 45,809 | 14 | 91,619 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Alyona is celebrating Happy Birthday! Her mother has an array of n flowers. Each flower has some mood, the mood of i-th flower is ai. The mood can be positive, zero or negative.
Let's define a subarray as a segment of consecutive flowers. The mother suggested some set of subarrays. Alyona wants to choose several of the subarrays suggested by her mother. After that, each of the flowers will add to the girl's happiness its mood multiplied by the number of chosen subarrays the flower is in.
For example, consider the case when the mother has 5 flowers, and their moods are equal to 1, - 2, 1, 3, - 4. Suppose the mother suggested subarrays (1, - 2), (3, - 4), (1, 3), (1, - 2, 1, 3). Then if the girl chooses the third and the fourth subarrays then:
* the first flower adds 1·1 = 1 to the girl's happiness, because he is in one of chosen subarrays,
* the second flower adds ( - 2)·1 = - 2, because he is in one of chosen subarrays,
* the third flower adds 1·2 = 2, because he is in two of chosen subarrays,
* the fourth flower adds 3·2 = 6, because he is in two of chosen subarrays,
* the fifth flower adds ( - 4)·0 = 0, because he is in no chosen subarrays.
Thus, in total 1 + ( - 2) + 2 + 6 + 0 = 7 is added to the girl's happiness. Alyona wants to choose such subarrays from those suggested by the mother that the value added to her happiness would be as large as possible. Help her do this!
Alyona can choose any number of the subarrays, even 0 or all suggested by her mother.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100) — the number of flowers and the number of subarrays suggested by the mother.
The second line contains the flowers moods — n integers a1, a2, ..., an ( - 100 ≤ ai ≤ 100).
The next m lines contain the description of the subarrays suggested by the mother. The i-th of these lines contain two integers li and ri (1 ≤ li ≤ ri ≤ n) denoting the subarray a[li], a[li + 1], ..., a[ri].
Each subarray can encounter more than once.
Output
Print single integer — the maximum possible value added to the Alyona's happiness.
Examples
Input
5 4
1 -2 1 3 -4
1 2
4 5
3 4
1 4
Output
7
Input
4 3
1 2 3 4
1 3
2 4
1 1
Output
16
Input
2 2
-1 -2
1 1
1 2
Output
0
Note
The first example is the situation described in the statements.
In the second example Alyona should choose all subarrays.
The third example has answer 0 because Alyona can choose none of the subarrays. | instruction | 0 | 45,810 | 14 | 91,620 |
Tags: constructive algorithms
Correct Solution:
```
ii = lambda: int(input())
mi = lambda: map(int, input().split())
li = lambda: list(mi())
si = lambda: input()
n, m = mi()
l = li()
h = 0
for _ in range(m): x, y = mi(); h += max(0, sum((l[k] for k in range(x-1, y))))
print(h)
``` | output | 1 | 45,810 | 14 | 91,621 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Alyona is celebrating Happy Birthday! Her mother has an array of n flowers. Each flower has some mood, the mood of i-th flower is ai. The mood can be positive, zero or negative.
Let's define a subarray as a segment of consecutive flowers. The mother suggested some set of subarrays. Alyona wants to choose several of the subarrays suggested by her mother. After that, each of the flowers will add to the girl's happiness its mood multiplied by the number of chosen subarrays the flower is in.
For example, consider the case when the mother has 5 flowers, and their moods are equal to 1, - 2, 1, 3, - 4. Suppose the mother suggested subarrays (1, - 2), (3, - 4), (1, 3), (1, - 2, 1, 3). Then if the girl chooses the third and the fourth subarrays then:
* the first flower adds 1·1 = 1 to the girl's happiness, because he is in one of chosen subarrays,
* the second flower adds ( - 2)·1 = - 2, because he is in one of chosen subarrays,
* the third flower adds 1·2 = 2, because he is in two of chosen subarrays,
* the fourth flower adds 3·2 = 6, because he is in two of chosen subarrays,
* the fifth flower adds ( - 4)·0 = 0, because he is in no chosen subarrays.
Thus, in total 1 + ( - 2) + 2 + 6 + 0 = 7 is added to the girl's happiness. Alyona wants to choose such subarrays from those suggested by the mother that the value added to her happiness would be as large as possible. Help her do this!
Alyona can choose any number of the subarrays, even 0 or all suggested by her mother.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100) — the number of flowers and the number of subarrays suggested by the mother.
The second line contains the flowers moods — n integers a1, a2, ..., an ( - 100 ≤ ai ≤ 100).
The next m lines contain the description of the subarrays suggested by the mother. The i-th of these lines contain two integers li and ri (1 ≤ li ≤ ri ≤ n) denoting the subarray a[li], a[li + 1], ..., a[ri].
Each subarray can encounter more than once.
Output
Print single integer — the maximum possible value added to the Alyona's happiness.
Examples
Input
5 4
1 -2 1 3 -4
1 2
4 5
3 4
1 4
Output
7
Input
4 3
1 2 3 4
1 3
2 4
1 1
Output
16
Input
2 2
-1 -2
1 1
1 2
Output
0
Note
The first example is the situation described in the statements.
In the second example Alyona should choose all subarrays.
The third example has answer 0 because Alyona can choose none of the subarrays. | instruction | 0 | 45,811 | 14 | 91,622 |
Tags: constructive algorithms
Correct Solution:
```
n, m = map(int, input().split())
a = list(map(int, input().split()))
res = 0
for i in range(m):
curr = 0
l, r = map(int, input().split())
for j in range(l - 1, r):
curr += a[j]
if curr > 0:
res += curr
print(res)
``` | output | 1 | 45,811 | 14 | 91,623 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Alyona is celebrating Happy Birthday! Her mother has an array of n flowers. Each flower has some mood, the mood of i-th flower is ai. The mood can be positive, zero or negative.
Let's define a subarray as a segment of consecutive flowers. The mother suggested some set of subarrays. Alyona wants to choose several of the subarrays suggested by her mother. After that, each of the flowers will add to the girl's happiness its mood multiplied by the number of chosen subarrays the flower is in.
For example, consider the case when the mother has 5 flowers, and their moods are equal to 1, - 2, 1, 3, - 4. Suppose the mother suggested subarrays (1, - 2), (3, - 4), (1, 3), (1, - 2, 1, 3). Then if the girl chooses the third and the fourth subarrays then:
* the first flower adds 1·1 = 1 to the girl's happiness, because he is in one of chosen subarrays,
* the second flower adds ( - 2)·1 = - 2, because he is in one of chosen subarrays,
* the third flower adds 1·2 = 2, because he is in two of chosen subarrays,
* the fourth flower adds 3·2 = 6, because he is in two of chosen subarrays,
* the fifth flower adds ( - 4)·0 = 0, because he is in no chosen subarrays.
Thus, in total 1 + ( - 2) + 2 + 6 + 0 = 7 is added to the girl's happiness. Alyona wants to choose such subarrays from those suggested by the mother that the value added to her happiness would be as large as possible. Help her do this!
Alyona can choose any number of the subarrays, even 0 or all suggested by her mother.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100) — the number of flowers and the number of subarrays suggested by the mother.
The second line contains the flowers moods — n integers a1, a2, ..., an ( - 100 ≤ ai ≤ 100).
The next m lines contain the description of the subarrays suggested by the mother. The i-th of these lines contain two integers li and ri (1 ≤ li ≤ ri ≤ n) denoting the subarray a[li], a[li + 1], ..., a[ri].
Each subarray can encounter more than once.
Output
Print single integer — the maximum possible value added to the Alyona's happiness.
Examples
Input
5 4
1 -2 1 3 -4
1 2
4 5
3 4
1 4
Output
7
Input
4 3
1 2 3 4
1 3
2 4
1 1
Output
16
Input
2 2
-1 -2
1 1
1 2
Output
0
Note
The first example is the situation described in the statements.
In the second example Alyona should choose all subarrays.
The third example has answer 0 because Alyona can choose none of the subarrays.
Submitted Solution:
```
n,q=map(int,input().split())
a=list(map(int,input().split()))
for i in range(1,n):
a[i]=(a[i-1]+a[i])
ans=0
for i in range(q):
x,y=map(int,input().split())
temp=a[y-1]
if x!=1:
temp-=a[x-2]
ans=max(ans,ans+temp)
print(ans)
``` | instruction | 0 | 45,812 | 14 | 91,624 |
Yes | output | 1 | 45,812 | 14 | 91,625 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Alyona is celebrating Happy Birthday! Her mother has an array of n flowers. Each flower has some mood, the mood of i-th flower is ai. The mood can be positive, zero or negative.
Let's define a subarray as a segment of consecutive flowers. The mother suggested some set of subarrays. Alyona wants to choose several of the subarrays suggested by her mother. After that, each of the flowers will add to the girl's happiness its mood multiplied by the number of chosen subarrays the flower is in.
For example, consider the case when the mother has 5 flowers, and their moods are equal to 1, - 2, 1, 3, - 4. Suppose the mother suggested subarrays (1, - 2), (3, - 4), (1, 3), (1, - 2, 1, 3). Then if the girl chooses the third and the fourth subarrays then:
* the first flower adds 1·1 = 1 to the girl's happiness, because he is in one of chosen subarrays,
* the second flower adds ( - 2)·1 = - 2, because he is in one of chosen subarrays,
* the third flower adds 1·2 = 2, because he is in two of chosen subarrays,
* the fourth flower adds 3·2 = 6, because he is in two of chosen subarrays,
* the fifth flower adds ( - 4)·0 = 0, because he is in no chosen subarrays.
Thus, in total 1 + ( - 2) + 2 + 6 + 0 = 7 is added to the girl's happiness. Alyona wants to choose such subarrays from those suggested by the mother that the value added to her happiness would be as large as possible. Help her do this!
Alyona can choose any number of the subarrays, even 0 or all suggested by her mother.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100) — the number of flowers and the number of subarrays suggested by the mother.
The second line contains the flowers moods — n integers a1, a2, ..., an ( - 100 ≤ ai ≤ 100).
The next m lines contain the description of the subarrays suggested by the mother. The i-th of these lines contain two integers li and ri (1 ≤ li ≤ ri ≤ n) denoting the subarray a[li], a[li + 1], ..., a[ri].
Each subarray can encounter more than once.
Output
Print single integer — the maximum possible value added to the Alyona's happiness.
Examples
Input
5 4
1 -2 1 3 -4
1 2
4 5
3 4
1 4
Output
7
Input
4 3
1 2 3 4
1 3
2 4
1 1
Output
16
Input
2 2
-1 -2
1 1
1 2
Output
0
Note
The first example is the situation described in the statements.
In the second example Alyona should choose all subarrays.
The third example has answer 0 because Alyona can choose none of the subarrays.
Submitted Solution:
```
n, m = map(int, input().split())
data = list(map(int, input().split()))
ans = 0
for i in range(m):
li, ri = map(int, input().split())
li -= 1
ans += max(sum(data[li:ri]), 0)
print(ans)
``` | instruction | 0 | 45,813 | 14 | 91,626 |
Yes | output | 1 | 45,813 | 14 | 91,627 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Alyona is celebrating Happy Birthday! Her mother has an array of n flowers. Each flower has some mood, the mood of i-th flower is ai. The mood can be positive, zero or negative.
Let's define a subarray as a segment of consecutive flowers. The mother suggested some set of subarrays. Alyona wants to choose several of the subarrays suggested by her mother. After that, each of the flowers will add to the girl's happiness its mood multiplied by the number of chosen subarrays the flower is in.
For example, consider the case when the mother has 5 flowers, and their moods are equal to 1, - 2, 1, 3, - 4. Suppose the mother suggested subarrays (1, - 2), (3, - 4), (1, 3), (1, - 2, 1, 3). Then if the girl chooses the third and the fourth subarrays then:
* the first flower adds 1·1 = 1 to the girl's happiness, because he is in one of chosen subarrays,
* the second flower adds ( - 2)·1 = - 2, because he is in one of chosen subarrays,
* the third flower adds 1·2 = 2, because he is in two of chosen subarrays,
* the fourth flower adds 3·2 = 6, because he is in two of chosen subarrays,
* the fifth flower adds ( - 4)·0 = 0, because he is in no chosen subarrays.
Thus, in total 1 + ( - 2) + 2 + 6 + 0 = 7 is added to the girl's happiness. Alyona wants to choose such subarrays from those suggested by the mother that the value added to her happiness would be as large as possible. Help her do this!
Alyona can choose any number of the subarrays, even 0 or all suggested by her mother.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100) — the number of flowers and the number of subarrays suggested by the mother.
The second line contains the flowers moods — n integers a1, a2, ..., an ( - 100 ≤ ai ≤ 100).
The next m lines contain the description of the subarrays suggested by the mother. The i-th of these lines contain two integers li and ri (1 ≤ li ≤ ri ≤ n) denoting the subarray a[li], a[li + 1], ..., a[ri].
Each subarray can encounter more than once.
Output
Print single integer — the maximum possible value added to the Alyona's happiness.
Examples
Input
5 4
1 -2 1 3 -4
1 2
4 5
3 4
1 4
Output
7
Input
4 3
1 2 3 4
1 3
2 4
1 1
Output
16
Input
2 2
-1 -2
1 1
1 2
Output
0
Note
The first example is the situation described in the statements.
In the second example Alyona should choose all subarrays.
The third example has answer 0 because Alyona can choose none of the subarrays.
Submitted Solution:
```
I = lambda: map(int, input().split())
n, m = I()
a = list(I())
result = 0
for _ in range(m):
l, r = I()
result += max(sum(a[l-1: r]), 0)
print(result)
``` | instruction | 0 | 45,814 | 14 | 91,628 |
Yes | output | 1 | 45,814 | 14 | 91,629 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Alyona is celebrating Happy Birthday! Her mother has an array of n flowers. Each flower has some mood, the mood of i-th flower is ai. The mood can be positive, zero or negative.
Let's define a subarray as a segment of consecutive flowers. The mother suggested some set of subarrays. Alyona wants to choose several of the subarrays suggested by her mother. After that, each of the flowers will add to the girl's happiness its mood multiplied by the number of chosen subarrays the flower is in.
For example, consider the case when the mother has 5 flowers, and their moods are equal to 1, - 2, 1, 3, - 4. Suppose the mother suggested subarrays (1, - 2), (3, - 4), (1, 3), (1, - 2, 1, 3). Then if the girl chooses the third and the fourth subarrays then:
* the first flower adds 1·1 = 1 to the girl's happiness, because he is in one of chosen subarrays,
* the second flower adds ( - 2)·1 = - 2, because he is in one of chosen subarrays,
* the third flower adds 1·2 = 2, because he is in two of chosen subarrays,
* the fourth flower adds 3·2 = 6, because he is in two of chosen subarrays,
* the fifth flower adds ( - 4)·0 = 0, because he is in no chosen subarrays.
Thus, in total 1 + ( - 2) + 2 + 6 + 0 = 7 is added to the girl's happiness. Alyona wants to choose such subarrays from those suggested by the mother that the value added to her happiness would be as large as possible. Help her do this!
Alyona can choose any number of the subarrays, even 0 or all suggested by her mother.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100) — the number of flowers and the number of subarrays suggested by the mother.
The second line contains the flowers moods — n integers a1, a2, ..., an ( - 100 ≤ ai ≤ 100).
The next m lines contain the description of the subarrays suggested by the mother. The i-th of these lines contain two integers li and ri (1 ≤ li ≤ ri ≤ n) denoting the subarray a[li], a[li + 1], ..., a[ri].
Each subarray can encounter more than once.
Output
Print single integer — the maximum possible value added to the Alyona's happiness.
Examples
Input
5 4
1 -2 1 3 -4
1 2
4 5
3 4
1 4
Output
7
Input
4 3
1 2 3 4
1 3
2 4
1 1
Output
16
Input
2 2
-1 -2
1 1
1 2
Output
0
Note
The first example is the situation described in the statements.
In the second example Alyona should choose all subarrays.
The third example has answer 0 because Alyona can choose none of the subarrays.
Submitted Solution:
```
p, q = map(int, input().split())
arr = [int(i) for i in input().split()]
ans = 0
b = [(list(map(int, input().split()))) for i in range(q)]
for i in range(q):
s = 0
for j in range(b[i][0] - 1, b[i][1]):
s += arr[j]
if s > 0:
ans += s
print(ans)
``` | instruction | 0 | 45,815 | 14 | 91,630 |
Yes | output | 1 | 45,815 | 14 | 91,631 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Alyona is celebrating Happy Birthday! Her mother has an array of n flowers. Each flower has some mood, the mood of i-th flower is ai. The mood can be positive, zero or negative.
Let's define a subarray as a segment of consecutive flowers. The mother suggested some set of subarrays. Alyona wants to choose several of the subarrays suggested by her mother. After that, each of the flowers will add to the girl's happiness its mood multiplied by the number of chosen subarrays the flower is in.
For example, consider the case when the mother has 5 flowers, and their moods are equal to 1, - 2, 1, 3, - 4. Suppose the mother suggested subarrays (1, - 2), (3, - 4), (1, 3), (1, - 2, 1, 3). Then if the girl chooses the third and the fourth subarrays then:
* the first flower adds 1·1 = 1 to the girl's happiness, because he is in one of chosen subarrays,
* the second flower adds ( - 2)·1 = - 2, because he is in one of chosen subarrays,
* the third flower adds 1·2 = 2, because he is in two of chosen subarrays,
* the fourth flower adds 3·2 = 6, because he is in two of chosen subarrays,
* the fifth flower adds ( - 4)·0 = 0, because he is in no chosen subarrays.
Thus, in total 1 + ( - 2) + 2 + 6 + 0 = 7 is added to the girl's happiness. Alyona wants to choose such subarrays from those suggested by the mother that the value added to her happiness would be as large as possible. Help her do this!
Alyona can choose any number of the subarrays, even 0 or all suggested by her mother.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100) — the number of flowers and the number of subarrays suggested by the mother.
The second line contains the flowers moods — n integers a1, a2, ..., an ( - 100 ≤ ai ≤ 100).
The next m lines contain the description of the subarrays suggested by the mother. The i-th of these lines contain two integers li and ri (1 ≤ li ≤ ri ≤ n) denoting the subarray a[li], a[li + 1], ..., a[ri].
Each subarray can encounter more than once.
Output
Print single integer — the maximum possible value added to the Alyona's happiness.
Examples
Input
5 4
1 -2 1 3 -4
1 2
4 5
3 4
1 4
Output
7
Input
4 3
1 2 3 4
1 3
2 4
1 1
Output
16
Input
2 2
-1 -2
1 1
1 2
Output
0
Note
The first example is the situation described in the statements.
In the second example Alyona should choose all subarrays.
The third example has answer 0 because Alyona can choose none of the subarrays.
Submitted Solution:
```
def findMood(arr, queries):
n = len(arr)
t = len(queries)
if n==0:
return 0
preSum = [arr[0]]
for i in range(1, n):
preSum.append(preSum[i-1]+arr[i])
maxm = float('-inf')
for i in range(t):
p = queries[i][0]-1
q = queries[i][1]-1
if p == 0:
currSum = preSum[q]
else:
currSum = preSum[q]-preSum[p-1]
if currSum+maxm > currSum:
maxm = currSum+maxm
elif currSum > maxm:
maxm = currSum
if maxm<0:
return 0
return maxm
li = list(map(int, input().split()))
arr = list(map(int, input().split()))
que = []
for i in range(li[1]):
que.append(list(map(int, input().split())))
print(findMood(arr, que))
``` | instruction | 0 | 45,816 | 14 | 91,632 |
No | output | 1 | 45,816 | 14 | 91,633 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Alyona is celebrating Happy Birthday! Her mother has an array of n flowers. Each flower has some mood, the mood of i-th flower is ai. The mood can be positive, zero or negative.
Let's define a subarray as a segment of consecutive flowers. The mother suggested some set of subarrays. Alyona wants to choose several of the subarrays suggested by her mother. After that, each of the flowers will add to the girl's happiness its mood multiplied by the number of chosen subarrays the flower is in.
For example, consider the case when the mother has 5 flowers, and their moods are equal to 1, - 2, 1, 3, - 4. Suppose the mother suggested subarrays (1, - 2), (3, - 4), (1, 3), (1, - 2, 1, 3). Then if the girl chooses the third and the fourth subarrays then:
* the first flower adds 1·1 = 1 to the girl's happiness, because he is in one of chosen subarrays,
* the second flower adds ( - 2)·1 = - 2, because he is in one of chosen subarrays,
* the third flower adds 1·2 = 2, because he is in two of chosen subarrays,
* the fourth flower adds 3·2 = 6, because he is in two of chosen subarrays,
* the fifth flower adds ( - 4)·0 = 0, because he is in no chosen subarrays.
Thus, in total 1 + ( - 2) + 2 + 6 + 0 = 7 is added to the girl's happiness. Alyona wants to choose such subarrays from those suggested by the mother that the value added to her happiness would be as large as possible. Help her do this!
Alyona can choose any number of the subarrays, even 0 or all suggested by her mother.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100) — the number of flowers and the number of subarrays suggested by the mother.
The second line contains the flowers moods — n integers a1, a2, ..., an ( - 100 ≤ ai ≤ 100).
The next m lines contain the description of the subarrays suggested by the mother. The i-th of these lines contain two integers li and ri (1 ≤ li ≤ ri ≤ n) denoting the subarray a[li], a[li + 1], ..., a[ri].
Each subarray can encounter more than once.
Output
Print single integer — the maximum possible value added to the Alyona's happiness.
Examples
Input
5 4
1 -2 1 3 -4
1 2
4 5
3 4
1 4
Output
7
Input
4 3
1 2 3 4
1 3
2 4
1 1
Output
16
Input
2 2
-1 -2
1 1
1 2
Output
0
Note
The first example is the situation described in the statements.
In the second example Alyona should choose all subarrays.
The third example has answer 0 because Alyona can choose none of the subarrays.
Submitted Solution:
```
n,m=[int(i) for i in input().split()]
a=[int(i) for i in input().split()]
sums=0
l=[]
for i in range(m):
b,c=[int(i) for i in input().split()]
l=a[b-1:c]
for i in range(0,len(l)):
sums=sums+(l[i]*l.count(l[i]))
if(sums<0):
sums=0
else:
l.append(sums)
print(sum(l))
``` | instruction | 0 | 45,817 | 14 | 91,634 |
No | output | 1 | 45,817 | 14 | 91,635 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Alyona is celebrating Happy Birthday! Her mother has an array of n flowers. Each flower has some mood, the mood of i-th flower is ai. The mood can be positive, zero or negative.
Let's define a subarray as a segment of consecutive flowers. The mother suggested some set of subarrays. Alyona wants to choose several of the subarrays suggested by her mother. After that, each of the flowers will add to the girl's happiness its mood multiplied by the number of chosen subarrays the flower is in.
For example, consider the case when the mother has 5 flowers, and their moods are equal to 1, - 2, 1, 3, - 4. Suppose the mother suggested subarrays (1, - 2), (3, - 4), (1, 3), (1, - 2, 1, 3). Then if the girl chooses the third and the fourth subarrays then:
* the first flower adds 1·1 = 1 to the girl's happiness, because he is in one of chosen subarrays,
* the second flower adds ( - 2)·1 = - 2, because he is in one of chosen subarrays,
* the third flower adds 1·2 = 2, because he is in two of chosen subarrays,
* the fourth flower adds 3·2 = 6, because he is in two of chosen subarrays,
* the fifth flower adds ( - 4)·0 = 0, because he is in no chosen subarrays.
Thus, in total 1 + ( - 2) + 2 + 6 + 0 = 7 is added to the girl's happiness. Alyona wants to choose such subarrays from those suggested by the mother that the value added to her happiness would be as large as possible. Help her do this!
Alyona can choose any number of the subarrays, even 0 or all suggested by her mother.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100) — the number of flowers and the number of subarrays suggested by the mother.
The second line contains the flowers moods — n integers a1, a2, ..., an ( - 100 ≤ ai ≤ 100).
The next m lines contain the description of the subarrays suggested by the mother. The i-th of these lines contain two integers li and ri (1 ≤ li ≤ ri ≤ n) denoting the subarray a[li], a[li + 1], ..., a[ri].
Each subarray can encounter more than once.
Output
Print single integer — the maximum possible value added to the Alyona's happiness.
Examples
Input
5 4
1 -2 1 3 -4
1 2
4 5
3 4
1 4
Output
7
Input
4 3
1 2 3 4
1 3
2 4
1 1
Output
16
Input
2 2
-1 -2
1 1
1 2
Output
0
Note
The first example is the situation described in the statements.
In the second example Alyona should choose all subarrays.
The third example has answer 0 because Alyona can choose none of the subarrays.
Submitted Solution:
```
def main():
n,m=[int(i) for i in input().split(' ')]
a=[0]+[int(i) for i in input().split(' ')]
lr=[]
for i in range(m):
li,ri=[int(i) for i in input().split(' ')]
lr.append(sum(a[li:ri+1]))
if max(a)<=0:
print(0)
else:
s,ms=0,0
for i in range(len(lr)):
s+=lr[i]
if s>ms: ms=s
elif s<0:s=0
return ms
print(main())
``` | instruction | 0 | 45,818 | 14 | 91,636 |
No | output | 1 | 45,818 | 14 | 91,637 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Alyona is celebrating Happy Birthday! Her mother has an array of n flowers. Each flower has some mood, the mood of i-th flower is ai. The mood can be positive, zero or negative.
Let's define a subarray as a segment of consecutive flowers. The mother suggested some set of subarrays. Alyona wants to choose several of the subarrays suggested by her mother. After that, each of the flowers will add to the girl's happiness its mood multiplied by the number of chosen subarrays the flower is in.
For example, consider the case when the mother has 5 flowers, and their moods are equal to 1, - 2, 1, 3, - 4. Suppose the mother suggested subarrays (1, - 2), (3, - 4), (1, 3), (1, - 2, 1, 3). Then if the girl chooses the third and the fourth subarrays then:
* the first flower adds 1·1 = 1 to the girl's happiness, because he is in one of chosen subarrays,
* the second flower adds ( - 2)·1 = - 2, because he is in one of chosen subarrays,
* the third flower adds 1·2 = 2, because he is in two of chosen subarrays,
* the fourth flower adds 3·2 = 6, because he is in two of chosen subarrays,
* the fifth flower adds ( - 4)·0 = 0, because he is in no chosen subarrays.
Thus, in total 1 + ( - 2) + 2 + 6 + 0 = 7 is added to the girl's happiness. Alyona wants to choose such subarrays from those suggested by the mother that the value added to her happiness would be as large as possible. Help her do this!
Alyona can choose any number of the subarrays, even 0 or all suggested by her mother.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100) — the number of flowers and the number of subarrays suggested by the mother.
The second line contains the flowers moods — n integers a1, a2, ..., an ( - 100 ≤ ai ≤ 100).
The next m lines contain the description of the subarrays suggested by the mother. The i-th of these lines contain two integers li and ri (1 ≤ li ≤ ri ≤ n) denoting the subarray a[li], a[li + 1], ..., a[ri].
Each subarray can encounter more than once.
Output
Print single integer — the maximum possible value added to the Alyona's happiness.
Examples
Input
5 4
1 -2 1 3 -4
1 2
4 5
3 4
1 4
Output
7
Input
4 3
1 2 3 4
1 3
2 4
1 1
Output
16
Input
2 2
-1 -2
1 1
1 2
Output
0
Note
The first example is the situation described in the statements.
In the second example Alyona should choose all subarrays.
The third example has answer 0 because Alyona can choose none of the subarrays.
Submitted Solution:
```
n, m = map(int, input().split())
l = list(int(num) for num in input().strip().split())[:n]
print(l)
total_happiness = 0
while m!=0:
x, y = map(int, input().split())
total = 0
for i in range(x-1,y):
total+=l[i]
if total>0:
total_happiness+=total
m-=1
print(total_happiness)
``` | instruction | 0 | 45,819 | 14 | 91,638 |
No | output | 1 | 45,819 | 14 | 91,639 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Comrade Dujikov is busy choosing artists for Timofey's birthday and is recieving calls from Taymyr from Ilia-alpinist.
Ilia-alpinist calls every n minutes, i.e. in minutes n, 2n, 3n and so on. Artists come to the comrade every m minutes, i.e. in minutes m, 2m, 3m and so on. The day is z minutes long, i.e. the day consists of minutes 1, 2, ..., z. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute.
Input
The only string contains three integers — n, m and z (1 ≤ n, m, z ≤ 104).
Output
Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls.
Examples
Input
1 1 10
Output
10
Input
1 2 5
Output
2
Input
2 3 9
Output
1
Note
Taymyr is a place in the north of Russia.
In the first test the artists come each minute, as well as the calls, so we need to kill all of them.
In the second test we need to kill artists which come on the second and the fourth minutes.
In the third test — only the artist which comes on the sixth minute. | instruction | 0 | 45,820 | 14 | 91,640 |
Tags: brute force, implementation, math
Correct Solution:
```
import math
def lcm(a, b):
return a*b // math.gcd(a, b)
n, m, z = map(int, input().split())
print(z // lcm(n, m))
``` | output | 1 | 45,820 | 14 | 91,641 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Comrade Dujikov is busy choosing artists for Timofey's birthday and is recieving calls from Taymyr from Ilia-alpinist.
Ilia-alpinist calls every n minutes, i.e. in minutes n, 2n, 3n and so on. Artists come to the comrade every m minutes, i.e. in minutes m, 2m, 3m and so on. The day is z minutes long, i.e. the day consists of minutes 1, 2, ..., z. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute.
Input
The only string contains three integers — n, m and z (1 ≤ n, m, z ≤ 104).
Output
Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls.
Examples
Input
1 1 10
Output
10
Input
1 2 5
Output
2
Input
2 3 9
Output
1
Note
Taymyr is a place in the north of Russia.
In the first test the artists come each minute, as well as the calls, so we need to kill all of them.
In the second test we need to kill artists which come on the second and the fourth minutes.
In the third test — only the artist which comes on the sixth minute. | instruction | 0 | 45,821 | 14 | 91,642 |
Tags: brute force, implementation, math
Correct Solution:
```
line = input().split(sep=" ")
n= int(line[0])
m= int(line[1])
z= int(line[2])
count = 0
for i in range(max(n,m),z+1):
if(i %n == 0 and i%m == 0):
count+=1
print(count)
``` | output | 1 | 45,821 | 14 | 91,643 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Comrade Dujikov is busy choosing artists for Timofey's birthday and is recieving calls from Taymyr from Ilia-alpinist.
Ilia-alpinist calls every n minutes, i.e. in minutes n, 2n, 3n and so on. Artists come to the comrade every m minutes, i.e. in minutes m, 2m, 3m and so on. The day is z minutes long, i.e. the day consists of minutes 1, 2, ..., z. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute.
Input
The only string contains three integers — n, m and z (1 ≤ n, m, z ≤ 104).
Output
Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls.
Examples
Input
1 1 10
Output
10
Input
1 2 5
Output
2
Input
2 3 9
Output
1
Note
Taymyr is a place in the north of Russia.
In the first test the artists come each minute, as well as the calls, so we need to kill all of them.
In the second test we need to kill artists which come on the second and the fourth minutes.
In the third test — only the artist which comes on the sixth minute. | instruction | 0 | 45,822 | 14 | 91,644 |
Tags: brute force, implementation, math
Correct Solution:
```
def gcd(a, b):
if a == 0:
return b
return gcd(b % a, a)
def lcm(a, b):
return (a * b) / gcd(a, b)
n,m,z=map(int,input().split())
ele=lcm(n,m)
print(int(z/ele))
``` | output | 1 | 45,822 | 14 | 91,645 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Comrade Dujikov is busy choosing artists for Timofey's birthday and is recieving calls from Taymyr from Ilia-alpinist.
Ilia-alpinist calls every n minutes, i.e. in minutes n, 2n, 3n and so on. Artists come to the comrade every m minutes, i.e. in minutes m, 2m, 3m and so on. The day is z minutes long, i.e. the day consists of minutes 1, 2, ..., z. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute.
Input
The only string contains three integers — n, m and z (1 ≤ n, m, z ≤ 104).
Output
Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls.
Examples
Input
1 1 10
Output
10
Input
1 2 5
Output
2
Input
2 3 9
Output
1
Note
Taymyr is a place in the north of Russia.
In the first test the artists come each minute, as well as the calls, so we need to kill all of them.
In the second test we need to kill artists which come on the second and the fourth minutes.
In the third test — only the artist which comes on the sixth minute. | instruction | 0 | 45,823 | 14 | 91,646 |
Tags: brute force, implementation, math
Correct Solution:
```
def gcd(x, y):
while y != 0:
x, y = y, x % y
return x
def lcm(x, y):
return (x * y) // gcd(x, y)
n, m, z = map(int, input().split())
print(z // lcm(n, m))
``` | output | 1 | 45,823 | 14 | 91,647 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Comrade Dujikov is busy choosing artists for Timofey's birthday and is recieving calls from Taymyr from Ilia-alpinist.
Ilia-alpinist calls every n minutes, i.e. in minutes n, 2n, 3n and so on. Artists come to the comrade every m minutes, i.e. in minutes m, 2m, 3m and so on. The day is z minutes long, i.e. the day consists of minutes 1, 2, ..., z. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute.
Input
The only string contains three integers — n, m and z (1 ≤ n, m, z ≤ 104).
Output
Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls.
Examples
Input
1 1 10
Output
10
Input
1 2 5
Output
2
Input
2 3 9
Output
1
Note
Taymyr is a place in the north of Russia.
In the first test the artists come each minute, as well as the calls, so we need to kill all of them.
In the second test we need to kill artists which come on the second and the fourth minutes.
In the third test — only the artist which comes on the sixth minute. | instruction | 0 | 45,824 | 14 | 91,648 |
Tags: brute force, implementation, math
Correct Solution:
```
# Description of the problem can be found at http://codeforces.com/problemset/problem/764/A
n, m, z = map(int, input().split())
t = 0
for i in range(1, z // m + 1):
if (i * m) % n == 0:
t += 1
print(t)
``` | output | 1 | 45,824 | 14 | 91,649 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Comrade Dujikov is busy choosing artists for Timofey's birthday and is recieving calls from Taymyr from Ilia-alpinist.
Ilia-alpinist calls every n minutes, i.e. in minutes n, 2n, 3n and so on. Artists come to the comrade every m minutes, i.e. in minutes m, 2m, 3m and so on. The day is z minutes long, i.e. the day consists of minutes 1, 2, ..., z. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute.
Input
The only string contains three integers — n, m and z (1 ≤ n, m, z ≤ 104).
Output
Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls.
Examples
Input
1 1 10
Output
10
Input
1 2 5
Output
2
Input
2 3 9
Output
1
Note
Taymyr is a place in the north of Russia.
In the first test the artists come each minute, as well as the calls, so we need to kill all of them.
In the second test we need to kill artists which come on the second and the fourth minutes.
In the third test — only the artist which comes on the sixth minute. | instruction | 0 | 45,825 | 14 | 91,650 |
Tags: brute force, implementation, math
Correct Solution:
```
n,m,z = [int(i) for i in input().split()]
count = 0
for i in range(1,z+1):
if (i%n == 0) and (i%m == 0):
count+=1
print(count)
``` | output | 1 | 45,825 | 14 | 91,651 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Comrade Dujikov is busy choosing artists for Timofey's birthday and is recieving calls from Taymyr from Ilia-alpinist.
Ilia-alpinist calls every n minutes, i.e. in minutes n, 2n, 3n and so on. Artists come to the comrade every m minutes, i.e. in minutes m, 2m, 3m and so on. The day is z minutes long, i.e. the day consists of minutes 1, 2, ..., z. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute.
Input
The only string contains three integers — n, m and z (1 ≤ n, m, z ≤ 104).
Output
Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls.
Examples
Input
1 1 10
Output
10
Input
1 2 5
Output
2
Input
2 3 9
Output
1
Note
Taymyr is a place in the north of Russia.
In the first test the artists come each minute, as well as the calls, so we need to kill all of them.
In the second test we need to kill artists which come on the second and the fourth minutes.
In the third test — only the artist which comes on the sixth minute. | instruction | 0 | 45,826 | 14 | 91,652 |
Tags: brute force, implementation, math
Correct Solution:
```
def solve(n, m, z):
x = set(range(n, z+1, n))
y = set(range(m, z+1, m))
z = list(range(1, z+1))
c = 0
for i in z:
if i in x and i in y:
c += 1
return c
def main():
n, m, z = list(map(int, input().split()))
print(solve(n, m, z))
main()
``` | output | 1 | 45,826 | 14 | 91,653 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Comrade Dujikov is busy choosing artists for Timofey's birthday and is recieving calls from Taymyr from Ilia-alpinist.
Ilia-alpinist calls every n minutes, i.e. in minutes n, 2n, 3n and so on. Artists come to the comrade every m minutes, i.e. in minutes m, 2m, 3m and so on. The day is z minutes long, i.e. the day consists of minutes 1, 2, ..., z. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute.
Input
The only string contains three integers — n, m and z (1 ≤ n, m, z ≤ 104).
Output
Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls.
Examples
Input
1 1 10
Output
10
Input
1 2 5
Output
2
Input
2 3 9
Output
1
Note
Taymyr is a place in the north of Russia.
In the first test the artists come each minute, as well as the calls, so we need to kill all of them.
In the second test we need to kill artists which come on the second and the fourth minutes.
In the third test — only the artist which comes on the sixth minute. | instruction | 0 | 45,827 | 14 | 91,654 |
Tags: brute force, implementation, math
Correct Solution:
```
#Taymyr is calling you
n,m,z = map(int,input().split())
c = 0
li1,li2 = [],[]
for i in range(n,z+1,n):
li1.append(i)
for i in range(m,z+1,m):
li2.append(i)
for i in li2:
if i in li1:
c+=1
print(c)
``` | output | 1 | 45,827 | 14 | 91,655 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Comrade Dujikov is busy choosing artists for Timofey's birthday and is recieving calls from Taymyr from Ilia-alpinist.
Ilia-alpinist calls every n minutes, i.e. in minutes n, 2n, 3n and so on. Artists come to the comrade every m minutes, i.e. in minutes m, 2m, 3m and so on. The day is z minutes long, i.e. the day consists of minutes 1, 2, ..., z. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute.
Input
The only string contains three integers — n, m and z (1 ≤ n, m, z ≤ 104).
Output
Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls.
Examples
Input
1 1 10
Output
10
Input
1 2 5
Output
2
Input
2 3 9
Output
1
Note
Taymyr is a place in the north of Russia.
In the first test the artists come each minute, as well as the calls, so we need to kill all of them.
In the second test we need to kill artists which come on the second and the fourth minutes.
In the third test — only the artist which comes on the sixth minute.
Submitted Solution:
```
text = input().split()
n = int(text[0])
m = int(text[1])
z = int(text[2])
a = n
b = m
while a != 0:
na = b % a
nb = a
a = na
b = nb
nok = int((n * m)/b)
otv = int(z/nok)
print(otv)
``` | instruction | 0 | 45,828 | 14 | 91,656 |
Yes | output | 1 | 45,828 | 14 | 91,657 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Comrade Dujikov is busy choosing artists for Timofey's birthday and is recieving calls from Taymyr from Ilia-alpinist.
Ilia-alpinist calls every n minutes, i.e. in minutes n, 2n, 3n and so on. Artists come to the comrade every m minutes, i.e. in minutes m, 2m, 3m and so on. The day is z minutes long, i.e. the day consists of minutes 1, 2, ..., z. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute.
Input
The only string contains three integers — n, m and z (1 ≤ n, m, z ≤ 104).
Output
Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls.
Examples
Input
1 1 10
Output
10
Input
1 2 5
Output
2
Input
2 3 9
Output
1
Note
Taymyr is a place in the north of Russia.
In the first test the artists come each minute, as well as the calls, so we need to kill all of them.
In the second test we need to kill artists which come on the second and the fourth minutes.
In the third test — only the artist which comes on the sixth minute.
Submitted Solution:
```
import math
nmz=[int(i) for i in input().split()]
n=nmz[0]
m=nmz[1]
z=nmz[2]
h=math.gcd(n,m)
l=int(n*m/h)
ans=int(z//l)
print(ans)
``` | instruction | 0 | 45,829 | 14 | 91,658 |
Yes | output | 1 | 45,829 | 14 | 91,659 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Comrade Dujikov is busy choosing artists for Timofey's birthday and is recieving calls from Taymyr from Ilia-alpinist.
Ilia-alpinist calls every n minutes, i.e. in minutes n, 2n, 3n and so on. Artists come to the comrade every m minutes, i.e. in minutes m, 2m, 3m and so on. The day is z minutes long, i.e. the day consists of minutes 1, 2, ..., z. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute.
Input
The only string contains three integers — n, m and z (1 ≤ n, m, z ≤ 104).
Output
Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls.
Examples
Input
1 1 10
Output
10
Input
1 2 5
Output
2
Input
2 3 9
Output
1
Note
Taymyr is a place in the north of Russia.
In the first test the artists come each minute, as well as the calls, so we need to kill all of them.
In the second test we need to kill artists which come on the second and the fourth minutes.
In the third test — only the artist which comes on the sixth minute.
Submitted Solution:
```
A = list(map(int, input().split()))
n = A[0]
m = A[1]
z = A[2]
k = 0
view = [i + 1 for i in range(z)]
for i in range(1, z + 1):
if i%m == 0: view[i - 1] = 'танцор'
if i%n == 0:
if view[i - 1] == 'танцор':
k+=1
view[i - 1] = 'звонок'
print(k)
``` | instruction | 0 | 45,830 | 14 | 91,660 |
Yes | output | 1 | 45,830 | 14 | 91,661 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Comrade Dujikov is busy choosing artists for Timofey's birthday and is recieving calls from Taymyr from Ilia-alpinist.
Ilia-alpinist calls every n minutes, i.e. in minutes n, 2n, 3n and so on. Artists come to the comrade every m minutes, i.e. in minutes m, 2m, 3m and so on. The day is z minutes long, i.e. the day consists of minutes 1, 2, ..., z. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute.
Input
The only string contains three integers — n, m and z (1 ≤ n, m, z ≤ 104).
Output
Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls.
Examples
Input
1 1 10
Output
10
Input
1 2 5
Output
2
Input
2 3 9
Output
1
Note
Taymyr is a place in the north of Russia.
In the first test the artists come each minute, as well as the calls, so we need to kill all of them.
In the second test we need to kill artists which come on the second and the fourth minutes.
In the third test — only the artist which comes on the sixth minute.
Submitted Solution:
```
z=input
n,m,k=(map(int,z().split()))
c=0
for i in range(1,k+1):
if i%n==00 and i%m==0 :
c+=1
print(c)
``` | instruction | 0 | 45,831 | 14 | 91,662 |
Yes | output | 1 | 45,831 | 14 | 91,663 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Comrade Dujikov is busy choosing artists for Timofey's birthday and is recieving calls from Taymyr from Ilia-alpinist.
Ilia-alpinist calls every n minutes, i.e. in minutes n, 2n, 3n and so on. Artists come to the comrade every m minutes, i.e. in minutes m, 2m, 3m and so on. The day is z minutes long, i.e. the day consists of minutes 1, 2, ..., z. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute.
Input
The only string contains three integers — n, m and z (1 ≤ n, m, z ≤ 104).
Output
Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls.
Examples
Input
1 1 10
Output
10
Input
1 2 5
Output
2
Input
2 3 9
Output
1
Note
Taymyr is a place in the north of Russia.
In the first test the artists come each minute, as well as the calls, so we need to kill all of them.
In the second test we need to kill artists which come on the second and the fourth minutes.
In the third test — only the artist which comes on the sixth minute.
Submitted Solution:
```
n,m,z = map(int,input().split())
i = 1
if m == 0:
print(0)
elif (n==m):
print(z//n)
else:
while(i<=z):
if(i%m == i%n and i%m == 0):
print(z//i)
i += 1
``` | instruction | 0 | 45,832 | 14 | 91,664 |
No | output | 1 | 45,832 | 14 | 91,665 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Comrade Dujikov is busy choosing artists for Timofey's birthday and is recieving calls from Taymyr from Ilia-alpinist.
Ilia-alpinist calls every n minutes, i.e. in minutes n, 2n, 3n and so on. Artists come to the comrade every m minutes, i.e. in minutes m, 2m, 3m and so on. The day is z minutes long, i.e. the day consists of minutes 1, 2, ..., z. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute.
Input
The only string contains three integers — n, m and z (1 ≤ n, m, z ≤ 104).
Output
Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls.
Examples
Input
1 1 10
Output
10
Input
1 2 5
Output
2
Input
2 3 9
Output
1
Note
Taymyr is a place in the north of Russia.
In the first test the artists come each minute, as well as the calls, so we need to kill all of them.
In the second test we need to kill artists which come on the second and the fourth minutes.
In the third test — only the artist which comes on the sixth minute.
Submitted Solution:
```
n, m, z = [int(x) for x in input().split()]
print(len(set(range(n, z+1, n)) and set(range(m, z+1, m))))
``` | instruction | 0 | 45,833 | 14 | 91,666 |
No | output | 1 | 45,833 | 14 | 91,667 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Comrade Dujikov is busy choosing artists for Timofey's birthday and is recieving calls from Taymyr from Ilia-alpinist.
Ilia-alpinist calls every n minutes, i.e. in minutes n, 2n, 3n and so on. Artists come to the comrade every m minutes, i.e. in minutes m, 2m, 3m and so on. The day is z minutes long, i.e. the day consists of minutes 1, 2, ..., z. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute.
Input
The only string contains three integers — n, m and z (1 ≤ n, m, z ≤ 104).
Output
Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls.
Examples
Input
1 1 10
Output
10
Input
1 2 5
Output
2
Input
2 3 9
Output
1
Note
Taymyr is a place in the north of Russia.
In the first test the artists come each minute, as well as the calls, so we need to kill all of them.
In the second test we need to kill artists which come on the second and the fourth minutes.
In the third test — only the artist which comes on the sixth minute.
Submitted Solution:
```
n,m,z=map(int,input().split())
c=0
for i in range(1,z):
if i%n==0 and i%m==0:
c+=1
print(c)
``` | instruction | 0 | 45,834 | 14 | 91,668 |
No | output | 1 | 45,834 | 14 | 91,669 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Comrade Dujikov is busy choosing artists for Timofey's birthday and is recieving calls from Taymyr from Ilia-alpinist.
Ilia-alpinist calls every n minutes, i.e. in minutes n, 2n, 3n and so on. Artists come to the comrade every m minutes, i.e. in minutes m, 2m, 3m and so on. The day is z minutes long, i.e. the day consists of minutes 1, 2, ..., z. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute.
Input
The only string contains three integers — n, m and z (1 ≤ n, m, z ≤ 104).
Output
Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls.
Examples
Input
1 1 10
Output
10
Input
1 2 5
Output
2
Input
2 3 9
Output
1
Note
Taymyr is a place in the north of Russia.
In the first test the artists come each minute, as well as the calls, so we need to kill all of them.
In the second test we need to kill artists which come on the second and the fourth minutes.
In the third test — only the artist which comes on the sixth minute.
Submitted Solution:
```
n,m,z = map(int,input().split())
a= max(n,m)
b= min(n,m)
count=0
while a<=z:
if a%b==0:
count+=1
a+=a
else:
a+=a
print(count)
``` | instruction | 0 | 45,835 | 14 | 91,670 |
No | output | 1 | 45,835 | 14 | 91,671 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Arkady decides to observe a river for n consecutive days. The river's water level on each day is equal to some real value.
Arkady goes to the riverside each day and makes a mark on the side of the channel at the height of the water level, but if it coincides with a mark made before, no new mark is created. The water does not wash the marks away. Arkady writes down the number of marks strictly above the water level each day, on the i-th day this value is equal to mi.
Define di as the number of marks strictly under the water level on the i-th day. You are to find out the minimum possible sum of di over all days. There are no marks on the channel before the first day.
Input
The first line contains a single positive integer n (1 ≤ n ≤ 105) — the number of days.
The second line contains n space-separated integers m1, m2, ..., mn (0 ≤ mi < i) — the number of marks strictly above the water on each day.
Output
Output one single integer — the minimum possible sum of the number of marks strictly below the water level among all days.
Examples
Input
6
0 1 0 3 0 2
Output
6
Input
5
0 1 2 1 2
Output
1
Input
5
0 1 1 2 2
Output
0
Note
In the first example, the following figure shows an optimal case.
<image>
Note that on day 3, a new mark should be created because if not, there cannot be 3 marks above water on day 4. The total number of marks underwater is 0 + 0 + 2 + 0 + 3 + 1 = 6.
In the second example, the following figure shows an optimal case.
<image> | instruction | 0 | 45,876 | 14 | 91,752 |
Tags: data structures, dp, greedy
Correct Solution:
```
n = int(input())
m = list(map(int, input().split()))
starts = sorted([(i - m[i], i) for i in range(n)], reverse = True)
cur = answer = 0
for i in range(n):
while starts[-1][1] < i:
starts.pop()
cur = max(cur, i - starts[-1][0])
answer += cur - m[i]
print(answer)
``` | output | 1 | 45,876 | 14 | 91,753 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Arkady decides to observe a river for n consecutive days. The river's water level on each day is equal to some real value.
Arkady goes to the riverside each day and makes a mark on the side of the channel at the height of the water level, but if it coincides with a mark made before, no new mark is created. The water does not wash the marks away. Arkady writes down the number of marks strictly above the water level each day, on the i-th day this value is equal to mi.
Define di as the number of marks strictly under the water level on the i-th day. You are to find out the minimum possible sum of di over all days. There are no marks on the channel before the first day.
Input
The first line contains a single positive integer n (1 ≤ n ≤ 105) — the number of days.
The second line contains n space-separated integers m1, m2, ..., mn (0 ≤ mi < i) — the number of marks strictly above the water on each day.
Output
Output one single integer — the minimum possible sum of the number of marks strictly below the water level among all days.
Examples
Input
6
0 1 0 3 0 2
Output
6
Input
5
0 1 2 1 2
Output
1
Input
5
0 1 1 2 2
Output
0
Note
In the first example, the following figure shows an optimal case.
<image>
Note that on day 3, a new mark should be created because if not, there cannot be 3 marks above water on day 4. The total number of marks underwater is 0 + 0 + 2 + 0 + 3 + 1 = 6.
In the second example, the following figure shows an optimal case.
<image> | instruction | 0 | 45,877 | 14 | 91,754 |
Tags: data structures, dp, greedy
Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
ma = [1] * n
for i in range(1, n):
ma[i] = max(ma[i - 1], a[i] + 1)
for i in range(n - 2, -1, -1):
ma[i] = max(ma[i + 1] - 1, ma[i])
print(sum(ma) - sum(a) - n)
# Made By Mostafa_Khaled
``` | output | 1 | 45,877 | 14 | 91,755 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Arkady decides to observe a river for n consecutive days. The river's water level on each day is equal to some real value.
Arkady goes to the riverside each day and makes a mark on the side of the channel at the height of the water level, but if it coincides with a mark made before, no new mark is created. The water does not wash the marks away. Arkady writes down the number of marks strictly above the water level each day, on the i-th day this value is equal to mi.
Define di as the number of marks strictly under the water level on the i-th day. You are to find out the minimum possible sum of di over all days. There are no marks on the channel before the first day.
Input
The first line contains a single positive integer n (1 ≤ n ≤ 105) — the number of days.
The second line contains n space-separated integers m1, m2, ..., mn (0 ≤ mi < i) — the number of marks strictly above the water on each day.
Output
Output one single integer — the minimum possible sum of the number of marks strictly below the water level among all days.
Examples
Input
6
0 1 0 3 0 2
Output
6
Input
5
0 1 2 1 2
Output
1
Input
5
0 1 1 2 2
Output
0
Note
In the first example, the following figure shows an optimal case.
<image>
Note that on day 3, a new mark should be created because if not, there cannot be 3 marks above water on day 4. The total number of marks underwater is 0 + 0 + 2 + 0 + 3 + 1 = 6.
In the second example, the following figure shows an optimal case.
<image> | instruction | 0 | 45,878 | 14 | 91,756 |
Tags: data structures, dp, greedy
Correct Solution:
```
n=int(input())
a=list(map(int,input().split()))
b=[]
maxi=0
for i in range(n):
maxi=max(maxi,a[i]+1)
b.append(maxi)
c=[]
count=b[-1]
for i in range(n-1,-1,-1):
if count-1>=b[i]:
count-=1
c.append(count)
else:
c.append(count)
c=c[::-1]
ans=0
for i in range(n):
ans+=(c[i]-a[i]-1)
print(ans)
``` | output | 1 | 45,878 | 14 | 91,757 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Arkady decides to observe a river for n consecutive days. The river's water level on each day is equal to some real value.
Arkady goes to the riverside each day and makes a mark on the side of the channel at the height of the water level, but if it coincides with a mark made before, no new mark is created. The water does not wash the marks away. Arkady writes down the number of marks strictly above the water level each day, on the i-th day this value is equal to mi.
Define di as the number of marks strictly under the water level on the i-th day. You are to find out the minimum possible sum of di over all days. There are no marks on the channel before the first day.
Input
The first line contains a single positive integer n (1 ≤ n ≤ 105) — the number of days.
The second line contains n space-separated integers m1, m2, ..., mn (0 ≤ mi < i) — the number of marks strictly above the water on each day.
Output
Output one single integer — the minimum possible sum of the number of marks strictly below the water level among all days.
Examples
Input
6
0 1 0 3 0 2
Output
6
Input
5
0 1 2 1 2
Output
1
Input
5
0 1 1 2 2
Output
0
Note
In the first example, the following figure shows an optimal case.
<image>
Note that on day 3, a new mark should be created because if not, there cannot be 3 marks above water on day 4. The total number of marks underwater is 0 + 0 + 2 + 0 + 3 + 1 = 6.
In the second example, the following figure shows an optimal case.
<image> | instruction | 0 | 45,879 | 14 | 91,758 |
Tags: data structures, dp, greedy
Correct Solution:
```
n = int(input())
m = input().split()
t = []
for i in range(n):
m[i] = int(m[i])
if i == 0:
t.append(m[i]+1)
else:
t.append(max(t[i-1], m[i]+1))
s = t[n-1] - m[n-1] - 1
for i in range(n-2, -1, -1):
if t[i] < t[i+1]-1:
t[i] = t[i+1]-1
s += t[i] - m[i] - 1
print(s)
``` | output | 1 | 45,879 | 14 | 91,759 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Arkady decides to observe a river for n consecutive days. The river's water level on each day is equal to some real value.
Arkady goes to the riverside each day and makes a mark on the side of the channel at the height of the water level, but if it coincides with a mark made before, no new mark is created. The water does not wash the marks away. Arkady writes down the number of marks strictly above the water level each day, on the i-th day this value is equal to mi.
Define di as the number of marks strictly under the water level on the i-th day. You are to find out the minimum possible sum of di over all days. There are no marks on the channel before the first day.
Input
The first line contains a single positive integer n (1 ≤ n ≤ 105) — the number of days.
The second line contains n space-separated integers m1, m2, ..., mn (0 ≤ mi < i) — the number of marks strictly above the water on each day.
Output
Output one single integer — the minimum possible sum of the number of marks strictly below the water level among all days.
Examples
Input
6
0 1 0 3 0 2
Output
6
Input
5
0 1 2 1 2
Output
1
Input
5
0 1 1 2 2
Output
0
Note
In the first example, the following figure shows an optimal case.
<image>
Note that on day 3, a new mark should be created because if not, there cannot be 3 marks above water on day 4. The total number of marks underwater is 0 + 0 + 2 + 0 + 3 + 1 = 6.
In the second example, the following figure shows an optimal case.
<image> | instruction | 0 | 45,880 | 14 | 91,760 |
Tags: data structures, dp, greedy
Correct Solution:
```
import math,sys,bisect,heapq
from collections import defaultdict,Counter,deque
from itertools import groupby,accumulate
#sys.setrecursionlimit(200000000)
input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
ilele = lambda: map(int,input().split())
alele = lambda: list(map(int, input().split()))
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
#MOD = 1000000000 + 7
def Y(c): print(["NO","YES"][c])
def y(c): print(["no","yes"][c])
def Yy(c): print(["No","Yes"][c])
N = int(input())
A = alele()
B=[]
maxi = 0
for i in A:
maxi = max(maxi,i)
B.append(maxi)
for i in range(N-2,-1,-1):
if B[i+1] - B[i] > 1:
B[i] = B[i+1] - 1
#print(B)
Ans= 0
for i in range(N):
Ans += B[i]-A[i]
print(Ans)
``` | output | 1 | 45,880 | 14 | 91,761 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Arkady decides to observe a river for n consecutive days. The river's water level on each day is equal to some real value.
Arkady goes to the riverside each day and makes a mark on the side of the channel at the height of the water level, but if it coincides with a mark made before, no new mark is created. The water does not wash the marks away. Arkady writes down the number of marks strictly above the water level each day, on the i-th day this value is equal to mi.
Define di as the number of marks strictly under the water level on the i-th day. You are to find out the minimum possible sum of di over all days. There are no marks on the channel before the first day.
Input
The first line contains a single positive integer n (1 ≤ n ≤ 105) — the number of days.
The second line contains n space-separated integers m1, m2, ..., mn (0 ≤ mi < i) — the number of marks strictly above the water on each day.
Output
Output one single integer — the minimum possible sum of the number of marks strictly below the water level among all days.
Examples
Input
6
0 1 0 3 0 2
Output
6
Input
5
0 1 2 1 2
Output
1
Input
5
0 1 1 2 2
Output
0
Note
In the first example, the following figure shows an optimal case.
<image>
Note that on day 3, a new mark should be created because if not, there cannot be 3 marks above water on day 4. The total number of marks underwater is 0 + 0 + 2 + 0 + 3 + 1 = 6.
In the second example, the following figure shows an optimal case.
<image> | instruction | 0 | 45,881 | 14 | 91,762 |
Tags: data structures, dp, greedy
Correct Solution:
```
N = int(input())
above = list(map(int, input().split()))
if N == 1:
print(0)
quit()
required_mark = [0] * N
required_mark[N-2] = above[N-1]
for i in reversed(range(N-2)):
required_mark[i] = max(above[i+1], required_mark[i+1] - 1)
d = 0
mark = 1
for i in range(1, N):
if mark == above[i]:
mark += 1
elif mark >= required_mark[i]:
d += mark - above[i] - 1
else:
d += mark - above[i]
mark += 1
print(d)
``` | output | 1 | 45,881 | 14 | 91,763 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Arkady decides to observe a river for n consecutive days. The river's water level on each day is equal to some real value.
Arkady goes to the riverside each day and makes a mark on the side of the channel at the height of the water level, but if it coincides with a mark made before, no new mark is created. The water does not wash the marks away. Arkady writes down the number of marks strictly above the water level each day, on the i-th day this value is equal to mi.
Define di as the number of marks strictly under the water level on the i-th day. You are to find out the minimum possible sum of di over all days. There are no marks on the channel before the first day.
Input
The first line contains a single positive integer n (1 ≤ n ≤ 105) — the number of days.
The second line contains n space-separated integers m1, m2, ..., mn (0 ≤ mi < i) — the number of marks strictly above the water on each day.
Output
Output one single integer — the minimum possible sum of the number of marks strictly below the water level among all days.
Examples
Input
6
0 1 0 3 0 2
Output
6
Input
5
0 1 2 1 2
Output
1
Input
5
0 1 1 2 2
Output
0
Note
In the first example, the following figure shows an optimal case.
<image>
Note that on day 3, a new mark should be created because if not, there cannot be 3 marks above water on day 4. The total number of marks underwater is 0 + 0 + 2 + 0 + 3 + 1 = 6.
In the second example, the following figure shows an optimal case.
<image> | instruction | 0 | 45,882 | 14 | 91,764 |
Tags: data structures, dp, greedy
Correct Solution:
```
n = int(input())
l = list(map(int,input().split()))
dp = [1 for i in range(n)]
m=0
for i in range(1,n):
dp[i] = max(dp[i-1],l[i]+1)
for i in range(n-2,-1,-1):
dp[i] = max(dp[i],dp[i+1]-1)
ans=0
for i in range(n):
ans+=dp[i]-l[i]-1
print(ans)
``` | output | 1 | 45,882 | 14 | 91,765 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Arkady decides to observe a river for n consecutive days. The river's water level on each day is equal to some real value.
Arkady goes to the riverside each day and makes a mark on the side of the channel at the height of the water level, but if it coincides with a mark made before, no new mark is created. The water does not wash the marks away. Arkady writes down the number of marks strictly above the water level each day, on the i-th day this value is equal to mi.
Define di as the number of marks strictly under the water level on the i-th day. You are to find out the minimum possible sum of di over all days. There are no marks on the channel before the first day.
Input
The first line contains a single positive integer n (1 ≤ n ≤ 105) — the number of days.
The second line contains n space-separated integers m1, m2, ..., mn (0 ≤ mi < i) — the number of marks strictly above the water on each day.
Output
Output one single integer — the minimum possible sum of the number of marks strictly below the water level among all days.
Examples
Input
6
0 1 0 3 0 2
Output
6
Input
5
0 1 2 1 2
Output
1
Input
5
0 1 1 2 2
Output
0
Note
In the first example, the following figure shows an optimal case.
<image>
Note that on day 3, a new mark should be created because if not, there cannot be 3 marks above water on day 4. The total number of marks underwater is 0 + 0 + 2 + 0 + 3 + 1 = 6.
In the second example, the following figure shows an optimal case.
<image> | instruction | 0 | 45,883 | 14 | 91,766 |
Tags: data structures, dp, greedy
Correct Solution:
```
n = int(input())
above = list(map(int, input().split()))
total = [x + 1 for x in above]
# ensure at most increase by one
for i in range(0,n-1)[::-1]:
total[i] = max(total[i], total[i+1] - 1)
# ensure nondecreasing
for i in range(1,n):
total[i] = max(total[i], total[i-1])
below = [t - a - 1 for t, a in zip(total, above)]
#print("total =", total)
#print("above =", above)
#print("below =", below)
print(sum(below))
``` | output | 1 | 45,883 | 14 | 91,767 |
Provide tags and a correct Python 2 solution for this coding contest problem.
Arkady decides to observe a river for n consecutive days. The river's water level on each day is equal to some real value.
Arkady goes to the riverside each day and makes a mark on the side of the channel at the height of the water level, but if it coincides with a mark made before, no new mark is created. The water does not wash the marks away. Arkady writes down the number of marks strictly above the water level each day, on the i-th day this value is equal to mi.
Define di as the number of marks strictly under the water level on the i-th day. You are to find out the minimum possible sum of di over all days. There are no marks on the channel before the first day.
Input
The first line contains a single positive integer n (1 ≤ n ≤ 105) — the number of days.
The second line contains n space-separated integers m1, m2, ..., mn (0 ≤ mi < i) — the number of marks strictly above the water on each day.
Output
Output one single integer — the minimum possible sum of the number of marks strictly below the water level among all days.
Examples
Input
6
0 1 0 3 0 2
Output
6
Input
5
0 1 2 1 2
Output
1
Input
5
0 1 1 2 2
Output
0
Note
In the first example, the following figure shows an optimal case.
<image>
Note that on day 3, a new mark should be created because if not, there cannot be 3 marks above water on day 4. The total number of marks underwater is 0 + 0 + 2 + 0 + 3 + 1 = 6.
In the second example, the following figure shows an optimal case.
<image> | instruction | 0 | 45,884 | 14 | 91,768 |
Tags: data structures, dp, greedy
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
pr = stdout.write
def in_num():
return int(raw_input())
def in_arr():
return map(int,raw_input().split())
def pr_num(n):
stdout.write(str(n)+'\n')
def pr_arr(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return map(int,stdin.read().split())
range = xrange # not for python 3.0+
n=in_num()
l=in_arr()
arr=[0]*n
mx=0
ans=0
for i in range(n-1,-1,-1):
mx-=1
mx=max(0,mx)
mx=max(mx,l[i]+1)
arr[i]=mx
for i in range(n):
mx=max(mx,arr[i])
ans+=mx
ans-=(l[i]+1)
pr_num(ans)
``` | output | 1 | 45,884 | 14 | 91,769 |
Provide a correct Python 3 solution for this coding contest problem.
There are 2N people numbered 1 through 2N. The height of Person i is h_i.
How many ways are there to make N pairs of people such that the following conditions are satisfied? Compute the answer modulo 998,244,353.
* Each person is contained in exactly one pair.
* For each pair, the heights of the two people in the pair are different.
Two ways are considered different if for some p and q, Person p and Person q are paired in one way and not in the other.
Constraints
* 1 \leq N \leq 50,000
* 1 \leq h_i \leq 100,000
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
h_1
:
h_{2N}
Output
Print the answer.
Examples
Input
2
1
1
2
3
Output
2
Input
5
30
10
20
40
20
10
10
30
50
60
Output
516 | instruction | 0 | 45,931 | 14 | 91,862 |
"Correct Solution:
```
ROOT = 3
MOD = 998244353
roots = [pow(ROOT,(MOD-1)>>i,MOD) for i in range(24)] # 1 の 2^i 乗根
iroots = [pow(x,MOD-2,MOD) for x in roots] # 1 の 2^i 乗根の逆元
def untt(a,n):
for i in range(n):
m = 1<<(n-i-1)
for s in range(1<<i):
w_N = 1
s *= m*2
for p in range(m):
a[s+p], a[s+p+m] = (a[s+p]+a[s+p+m])%MOD, (a[s+p]-a[s+p+m])*w_N%MOD
w_N = w_N*roots[n-i]%MOD
def iuntt(a,n):
for i in range(n):
m = 1<<i
for s in range(1<<(n-i-1)):
w_N = 1
s *= m*2
for p in range(m):
a[s+p], a[s+p+m] = (a[s+p]+a[s+p+m]*w_N)%MOD, (a[s+p]-a[s+p+m]*w_N)%MOD
w_N = w_N*iroots[i+1]%MOD
inv = pow((MOD+1)//2,n,MOD)
for i in range(1<<n):
a[i] = a[i]*inv%MOD
def convolution(a,b):
la = len(a)
lb = len(b)
deg = la+lb-2
n = deg.bit_length()
if min(la, lb) <= 50:
if la < lb:
la,lb = lb,la
a,b = b,a
res = [0]*(la+lb-1)
for i in range(la):
for j in range(lb):
res[i+j] += a[i]*b[j]
res[i+j] %= MOD
return res
N = 1<<n
a += [0]*(N-len(a))
b += [0]*(N-len(b))
untt(a,n)
untt(b,n)
for i in range(N):
a[i] = a[i]*b[i]%MOD
iuntt(a,n)
return a[:deg+1]
def convolution_all(polys):
if not polys: return [1]
polys.sort(key=lambda x:len(x))
from collections import deque
q = deque(polys)
for _ in range(len(polys)-1):
a = q.popleft()
b = q.popleft()
q.append(convolution(a,b))
return q[0]
SIZE=10**5 + 10 #998244353 #ここを変更する
inv = [0]*SIZE # inv[j] = j^{-1} mod MOD
fac = [0]*SIZE # fac[j] = j! mod MOD
finv = [0]*SIZE # finv[j] = (j!)^{-1} mod MOD
fac[0] = fac[1] = 1
finv[0] = finv[1] = 1
for i in range(2,SIZE):
fac[i] = fac[i-1]*i%MOD
finv[-1] = pow(fac[-1],MOD-2,MOD)
for i in range(SIZE-1,0,-1):
finv[i-1] = finv[i]*i%MOD
inv[i] = finv[i]*fac[i-1]%MOD
def choose(n,r): # nCk mod MOD の計算
if 0 <= r <= n:
return (fac[n]*finv[r]%MOD)*finv[n-r]%MOD
else:
return 0
dfac = [0]*SIZE # fac[j] = j! mod MOD
dfac[0] = dfac[1] = 1
for i in range(2,SIZE):
dfac[i] = dfac[i-2]*i%MOD
import sys
readline = sys.stdin.readline
read = sys.stdin.read
n,*h = map(int, read().split())
from collections import Counter
def get(k): # k 個の頂点からiペア作れる場合の数
res = [1]
for i in range(k//2):
res.append(res[-1]*choose(k-2*i,2)%MOD)
for i in range(2,k//2 + 1):
res[i] = res[i]*finv[i]%MOD
return res
d = Counter(h)
res = []
for k,v in d.items():
if v >= 2: res.append(get(v))
if v > n:
print(0)
exit()
bad = convolution_all(res)
#print(bad)
#bad = [1,5,7,3]
ans = 0
sgn = 1
for i,ai in enumerate(bad):
ans += sgn*(dfac[2*n-2*i-1] if i < n else 1)*ai
ans %= MOD
sgn *= -1
#print(ans)
print(ans%MOD)
``` | output | 1 | 45,931 | 14 | 91,863 |
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