TnT/CodeNet4Repair huggingface dataset:
Problem:
Quickly after finishing the tutorial of the online game _ATChat_ , you have decided to visit a particular place with N-1 players who happen to be there. These N players, including you, are numbered 1 through N, and the **friendliness** of Player i is A_i.  The N players will arrive at the place one by one in some order. To make sure nobody gets lost, you have set the following rule: players who have already arrived there should form a circle, and a player who has just arrived there should cut into the circle somewhere.  When each player, except the first one to arrive, arrives at the place, the player gets **comfort** equal to the smaller of the friendliness of the clockwise adjacent player and that of the counter-clockwise adjacent player. The first player to arrive there gets the comfort of 0.  What is the maximum total comfort the N players can get by optimally choosing the order of arrivals and the positions in the circle to cut into?

Incorrect submission (verdict: Wrong Answer):
n = int(input())
c = list(map(int, input().split()))
c.sort(reverse = True)
a = 0
for i in range(int(n//2+1)):
  a += c[i]
if n%2==0:
  print(a*2-c[0])
else:
  print(a*2-c[0]-c[n//2])

Fixed, accepted submission:
n = int(input())
c = list(map(int, input().split()))
c.sort(reverse = True)
a = 0
for i in range(int((n+1)//2)):
  a += c[i]
if n%2==0:
  print(a*2-c[0])
else:
  print(a*2-c[0]-c[n//2])
Problem:
Takahashi, who works at DISCO, is standing before an iron bar. The bar has N-1 notches, which divide the bar into N sections. The i-th section from the left has a length of A_i millimeters.  Takahashi wanted to choose a notch and cut the bar at that point into two parts with the same length. However, this may not be possible as is, so he will do the following operations some number of times **before** he does the cut:    * Choose one section and expand it, increasing its length by 1 millimeter. Doing this operation once costs 1 yen (the currency of Japan).   * Choose one section of length at least 2 millimeters and shrink it, decreasing its length by 1 millimeter. Doing this operation once costs 1 yen.  Find the minimum amount of money needed before cutting the bar into two parts with the same length.

Incorrect submission (verdict: Wrong Answer):
n = int(input())
a= [int(x) for x in input().split(" ",n-1)]
print(a)
s = 0
for x in range(n):
    s += a[x-1]
r = 0
l = 0
x = 0
while r < s/2:
    r += a[x]
    x += 1
y = 1
while l < s/2:
    l += a[n-y]
    y += 1
if r == s//2:
    print(0)
else:
    k = min(r,l)  
    v = k - s/2
    m = v // 0.5
    print(int(m))

Fixed, accepted submission:
n = int(input())
a= [int(x) for x in input().split(" ",n-1)]
s = 0
for x in range(n):
    s += a[x-1]
r = 0
l = 0
x = 0
while r < s/2:
    r += a[x]
    x += 1
y = 1
while l < s/2:
    l += a[n-y]
    y += 1
if r == s//2:
    print(0)
else:
    k = min(r,l)  
    v = k - s/2
    m = v // 0.5
    print(int(m))
Problem:
You went shopping to buy cakes and donuts with X yen (the currency of Japan).  First, you bought one cake for A yen at a cake shop. Then, you bought as many donuts as possible for B yen each, at a donut shop.  How much do you have left after shopping?

Incorrect submission (verdict: Wrong Answer):
import sys

if sys.platform =='ios':
    sys.stdin=open('input_file.txt')
    
x,a,b=[int(input()) for i in range(3)]

print((x-a)//b)

Fixed, accepted submission:
import sys

if sys.platform =='ios':
    sys.stdin=open('input_file.txt')
    
x,a,b=[int(input()) for i in range(3)]

print((x-a)-((x-a)//b)*b)
Problem:
Write a program which calculates the cube of a given integer x.

Incorrect submission (verdict: Wrong Answer):
x = input() 
n = (int(x)**3)
print("n =" ,n)


Fixed, accepted submission:
x = input() 
n = (int(x)**3)
print(n)
Problem:
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A.  The problem has N test cases.  For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively.  See the Output section for the output format.

Incorrect submission (verdict: Wrong Answer):
def main():
    N = int(input())
    ac = 0
    wa = 0
    tle = 0
    re = 0
    for i in range(0, N):
        res = input()
        if (res == 'AC'):
            ac += 1
        if (res == 'WA'):
            wa += 1
        if (res == 'TLE'):
            tle += 1
        if (res == 'RE'):
            re += 1
    print(f"AC x {ac}")
    print(f"WA x {wa}")
    print(f"TEL x {tle}")
    print(f"RE x {re}")


if __name__ == '__main__':
    main()


Fixed, accepted submission:
def main():
    N = int(input())
    ac = 0
    wa = 0
    tle = 0
    re = 0
    for i in range(0, N):
        res = input()
        if (res == 'AC'):
            ac += 1
        if (res == 'WA'):
            wa += 1
        if (res == 'TLE'):
            tle += 1
        if (res == 'RE'):
            re += 1
    print(f"AC x {ac}")
    print(f"WA x {wa}")
    print(f"TLE x {tle}")
    print(f"RE x {re}")


if __name__ == '__main__':
    main()
Problem:
There are four towns, numbered 1,2,3 and 4. Also, there are three roads. The i-th road connects different towns a_i and b_i bidirectionally. No two roads connect the same pair of towns. Other than these roads, there is no way to travel between these towns, but any town can be reached from any other town using these roads.  Determine if we can visit all the towns by traversing each of the roads exactly once.

Incorrect submission (verdict: Wrong Answer):
l = []

for i in range(3):
    l += list(map(int, input().split()))
print(l)

for i in range(4):
    if l.count(i) == 3:
        print('NO')
        break
else:
    print('YES')

Fixed, accepted submission:
l = []

for _ in range(3):
    l += list(map(int, input().split()))

for i in range(4):
    if l.count(i + 1) == 3:
        print('NO')
        break
else:
    print('YES')
Problem:
We have a sequence of length N, a = (a_1, a_2, ..., a_N). Each a_i is a positive integer.  Snuke's objective is to permute the element in a so that the following condition is satisfied:    * For each 1 ≤ i ≤ N - 1, the product of a_i and a_{i + 1} is a multiple of 4.  Determine whether Snuke can achieve his objective.

Incorrect submission (verdict: Wrong Answer):
n = int(input())
a_list = list(map(int,input().split()))
zero = 0
one  = 0
two = 0

for a in a_list:
    remain = a%2
    if remain == 0:
        if a/2 >= 2:
            two += 1
        elif a/2 == 1:
            one += 1
    else:
        zero += 1

if one:
    zero += 1

if zero <= two+1 or zero <= two:
    print('yes')
else:
    print('no')


Fixed, accepted submission:
n = int(input())
a_list = list(map(int,input().split()))

divs = [a%4 for a in a_list]
mp_of_four = divs.count(0)
mp_of_two  = divs.count(2) 

if mp_of_four+mp_of_two//2 >= n//2:
    print('Yes')
else:
    print('No')
Problem:
Sig has built his own keyboard. Designed for ultimate simplicity, this keyboard only has 3 keys on it: the `0` key, the `1` key and the backspace key.  To begin with, he is using a plain text editor with this keyboard. This editor always displays one string (possibly empty). Just after the editor is launched, this string is empty. When each key on the keyboard is pressed, the following changes occur to the string:    * The `0` key: a letter `0` will be inserted to the right of the string.   * The `1` key: a letter `1` will be inserted to the right of the string.   * The backspace key: if the string is empty, nothing happens. Otherwise, the rightmost letter of the string is deleted.  Sig has launched the editor, and pressed these keys several times. You are given a string s, which is a record of his keystrokes in order. In this string, the letter `0` stands for the `0` key, the letter `1` stands for the `1` key and the letter `B` stands for the backspace key. What string is displayed in the editor now?

Incorrect submission (verdict: Wrong Answer):
s=input()
a=''
for i in s:
    if i=='B':
        if len(a)!=0:a=a[:-1]
    else:a+=i
    print(a,i)
print(a)

Fixed, accepted submission:
a=''
for i in input():
    if i=='B':
        if len(a)!=0:a=a[:-1]
    else:a+=i
print(a)
Problem:
There is a rooted tree (see Notes) with N vertices numbered 1 to N. Each of the vertices, except the root, has a directed edge coming from its parent. Note that the root may not be Vertex 1.  Takahashi has added M new directed edges to this graph. Each of these M edges, u \rightarrow v, extends from some vertex u to its descendant v.  You are given the directed graph with N vertices and N-1+M edges after Takahashi added edges. More specifically, you are given N-1+M pairs of integers, (A_1, B_1), ..., (A_{N-1+M}, B_{N-1+M}), which represent that the i-th edge extends from Vertex A_i to Vertex B_i.  Restore the original rooted tree.

Incorrect submission (verdict: Wrong Answer):
from collections import deque
n,m = map(int,input().split())
e = [[] for i in range(n)]
d = [1]*n
s = [0]*n
for i in range(n+m-1):
    a,b = map(int,input().split())
    e[a-1].append(b-1)
    s[b-1] += 1

if 0 in s:
    ans = [-1]*n
    p = s.index(0)
    ans[p] = 0
    q = deque([])
    q.append(p)
    while q:
        now = q.popleft()
        for nex in e[now]:
            if ans[nex] == -1:
                ans[nex] = now+1
                q.append(nex)
    for i in ans:
        print(i)
    exit()




Fixed, accepted submission:
from collections import deque
n,m = map(int,input().split())
e = [[] for i in range(n)]
d = [1]*n
s = [0]*n
t = [[] for i in range(n)]
for i in range(n+m-1):
    a,b = map(int,input().split())
    a -= 1
    b -= 1
    e[a].append(b)
    t[b].append(a)
    s[b] += 1


ans = [-1]*n
p = s.index(0)
ans[p] = 0
q = deque([])
q.append(p)
s = set()

while q:
    now = q.popleft()
    s.add(now)
    for nex in e[now]:
        if ans[nex] == -1:
            check = True
            for j in t[nex]:
                if j not in s:
                    check = False
                    break
            if check:
                ans[nex] = now+1
                q.append(nex)
for i in ans:
    print(i)
Problem:
Write a program which calculates the cube of a given integer x.

Incorrect submission (verdict: Wrong Answer):
def f(x):
    return x**3 

Fixed, accepted submission:
def cube(x):
    return x**3

def main():
    x = int(input())
    ans = cube(x)
    print(ans)

if __name__=='__main__':
    main()
Problem:
When a boy was cleaning up after his grand father passing, he found an old paper:         In addition, other side of the paper says that "go ahead a number of steps equivalent to the first integer, and turn clockwise by degrees equivalent to the second integer".  His grand mother says that Sanbonmatsu was standing at the center of town. However, now buildings are crammed side by side and people can not walk along exactly what the paper says in. Your task is to write a program which hunts for the treature on the paper.  For simplicity, 1 step is equivalent to 1 meter. Input consists of several pairs of two integers d (the first integer) and t (the second integer) separated by a comma. Input ends with "0, 0". Your program should print the coordinate (x, y) of the end point. There is the treature where x meters to the east and y meters to the north from the center of town.  You can assume that d ≤ 100 and -180 ≤ t ≤ 180\.

Incorrect submission (verdict: Wrong Answer):
from math import radians
from cmath import rect

z = 0
deg = 90
while True:
    r, d = map(int, input().split(','))
    if r == d == 0: break
    z += rect(r, radians(deg))
    deg += d
print(-int(z.real), int(z.imag))

Fixed, accepted submission:
from math import radians
from cmath import rect

z = 0
deg = 90
while True:
    r, d = map(float, input().split(','))
    if r == d == 0: break
    z += rect(r, radians(deg))
    deg -= d
print(int(z.real))
print(int(z.imag))
Problem:
The problem set at _CODE FESTIVAL 20XX Finals_ consists of N problems.  The score allocated to the i-th (1≦i≦N) problem is i points.  Takahashi, a contestant, is trying to score exactly N points. For that, he is deciding which problems to solve.  As problems with higher scores are harder, he wants to minimize the highest score of a problem among the ones solved by him.  Determine the set of problems that should be solved.

Incorrect submission (verdict: Wrong Answer):
# -*- coding: utf-8 -*-
"""
Created on Fri Jun 12 20:20:11 2020

@author: NEC-PCuser
"""


N = int(input())

i = 1
li = []
while (N >= (i * (i + 1)) // 2):
    li.append(i)
    i += 1
    

value = (i - 1) * i // 2
print(value)

index = len(li) - 1
for i in range(value, N):
    li[index] += 1
    index -= 1

for i in li:
    print(i)

Fixed, accepted submission:
# -*- coding: utf-8 -*-
"""
Created on Fri Jun 12 20:20:11 2020

@author: NEC-PCuser
"""


N = int(input())

i = 1
li = []
while (N >= (i * (i + 1)) // 2):
    li.append(i)
    i += 1
    

value = (i - 1) * i // 2

index = len(li) - 1
for i in range(value, N):
    li[index] += 1
    index -= 1

for i in li:
    print(i)
Problem:
You are given three strings A, B and C. Check whether they form a _word chain_.  More formally, determine whether both of the following are true:    * The last character in A and the initial character in B are the same.   * The last character in B and the initial character in C are the same.  If both are true, print `YES`. Otherwise, print `NO`.

Incorrect submission (verdict: Wrong Answer):
A,B,C = input().split()

X = False
if A[-1]==B[0]:
    if B[-1] == C[0]:
        X = True

if X:
    print('Yes')
else:
    print('No')



Fixed, accepted submission:
A,B,C = input().split()

X = False
if A[-1]==B[0]:
    if B[-1] == C[0]:
        X = True

if X:
    print('YES')
else:
    print('NO')
Problem:
Snuke and Raccoon have a heap of N cards. The i-th card from the top has the integer a_i written on it.  They will share these cards. First, Snuke will take some number of cards from the top of the heap, then Raccoon will take all the remaining cards. Here, both Snuke and Raccoon have to take at least one card.  Let the sum of the integers on Snuke's cards and Raccoon's cards be x and y, respectively. They would like to minimize |x-y|. Find the minimum possible value of |x-y|.

Incorrect submission (verdict: Wrong Answer):
n=int(input())
arr=list(map(int,input().split()))
for i in range(n-1):
  arr[i+1]+=arr[i]
print(arr)
ans=10**18
for i in range(n-1):
  x=arr[i]
  y=arr[n-1]-x
  tmp=abs(x-y)
  ans=min(ans,tmp)
print(ans)

Fixed, accepted submission:
n=int(input())
arr=list(map(int,input().split()))
for i in range(n-1):
  arr[i+1]+=arr[i]
ans=10**18
for i in range(n-1):
  x=arr[i]
  y=arr[n-1]-x
  tmp=abs(x-y)
  ans=min(ans,tmp)
print(ans)
Problem:
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?

Incorrect submission (verdict: Wrong Answer):
n=int(input())
sum=0

for i in range(n):
  sum+=(n+1)**2
print(sum)


Fixed, accepted submission:
n=int(input())
sum=0

for i in range(n):
  sum+=(i+1)
print(sum)
Problem:
AtCoder Mart sells 1000000 of each of the six items below:    * Riceballs, priced at 100 yen (the currency of Japan) each   * Sandwiches, priced at 101 yen each   * Cookies, priced at 102 yen each   * Cakes, priced at 103 yen each   * Candies, priced at 104 yen each   * Computers, priced at 105 yen each  Takahashi wants to buy some of them that cost exactly X yen in total. Determine whether this is possible.   (Ignore consumption tax.)

Incorrect submission (verdict: Wrong Answer):
n = int(input())
if n <= 99:
  print('No')
  exit()
dp = [True] + [False] * n
for i in range(100, n + 1):
  if dp[i - 100] or dp[i - 101] or dp[i - 102] or dp[i - 103] or dp[i - 104] or dp[i - 105]:
    dp[i] = True
print(['No','Yes'][dp[n]])

Fixed, accepted submission:
n = int(input())
if n <= 99:
  print(0)
  exit()
if 100 <= n <= 105:
  print(1)
  exit()
dp = [True] + [False] * n
for i in range(100, n + 1):
  if dp[i - 100] or dp[i - 101] or dp[i - 102] or dp[i - 103] or dp[i - 104] or dp[i - 105]:
    dp[i] = True
print([0, 1][dp[n]])
Problem:
Write a program which reads two integers a, b and an operator op, and then prints the value of a op b.  The operator op is '+', '-', '*' or '/' (sum, difference, product or quotient). The division should truncate any fractional part.

Incorrect submission (verdict: Wrong Answer):
while True:
    a,op,b = input().split()
    a = int(a)
    b = int(b)
    if (op == "+"):
        print(a + b)
    elif(op == "-"):
        print(a + b)
    elif(op == "/"):
        print(a // b)
    elif(op == "*"):
        print(a + b)
    elif(op == "?"):
        break

Fixed, accepted submission:
while True:
    a,op,b = input().split()
    a = int(a)
    b = int(b)
    if (op == "+"):
        print(a + b)
    elif(op == "-"):
        print(a - b)
    elif(op == "/"):
        print(a // b)
    elif(op == "*"):
        print(a * b)
    elif(op == "?"):
        break
Problem:
Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive).

Incorrect submission (verdict: Time Limit Exceeded):
n, a, b = map(int, input().split())

def divide(num):
  rest = num
  sum = 0
  while rest > 0:
    num /= 10
    rest %= 10
    sum += rest
  return sum

for i in range(n):
  temp = divide(i+1)
  ans = 0
  if temp>= a and temp <= b:
    ans += temp
print(ans)

Fixed, accepted submission:
n, a, b = map(int, input().split())

def divide(num):
  sum = 0
  while num > 0:
    sum += num % 10
    num //= 10
  return sum

ans = 0
for i in range(n):
  temp = divide(i+1)
  if temp>= a and temp <= b:
    ans += (i+1)
print(ans)
Problem:
You are given an integer sequence of length N, a_1,a_2,...,a_N.  For each 1≤i≤N, you have three choices: add 1 to a_i, subtract 1 from a_i or do nothing.  After these operations, you select an integer X and count the number of i such that a_i=X.  Maximize this count by making optimal choices.

Incorrect submission (verdict: Wrong Answer):
from collections import defaultdict
N=int(input())
a=list(map(int,input().split()))
d=defaultdict(int)
for item in a:
    d[item-1]+=1
    d[item]+=1
    d[item+1]+=1
b=sorted(d.items(),key=lambda x:x[1],reverse=True)
print(b[0][0])

Fixed, accepted submission:
from collections import defaultdict
N=int(input())
a=list(map(int,input().split()))
d=defaultdict(int)
for item in a:
    d[item-1]+=1
    d[item]+=1
    d[item+1]+=1
b=sorted(d.items(),key=lambda x:x[1],reverse=True)
print(b[0][1])
Problem:
There is a cave.  The cave has N rooms and M passages. The rooms are numbered 1 to N, and the passages are numbered 1 to M. Passage i connects Room A_i and Room B_i bidirectionally. One can travel between any two rooms by traversing passages. Room 1 is a special room with an entrance from the outside.  It is dark in the cave, so we have decided to place a signpost in each room except Room 1. The signpost in each room will point to one of the rooms directly connected to that room with a passage.  Since it is dangerous in the cave, our objective is to satisfy the condition below for each room except Room 1.    * If you start in that room and repeatedly move to the room indicated by the signpost in the room you are in, you will reach Room 1 after traversing the minimum number of passages possible.  Determine whether there is a way to place signposts satisfying our objective, and print one such way if it exists.

Incorrect submission (verdict: Wrong Answer):
n, m = map(int, input().split())

graph = [[] for _ in range(n+1)]

for i in range(m):
    a, b = map(int, input().split())
    graph[a].append(b)
    graph[b].append(a)

# bfs
milestone = [0] * (n+1)
q = []
q.append(1)
while len(q) != 0:
    for i in graph[q[0]]:
        if milestone[i] == 0:
            milestone[i] = q[0]
            q.append(i)
    q.pop(0)

for i in range(2, n+1):
    print(milestone[i])

Fixed, accepted submission:
n, m = map(int, input().split())

graph = [[] for _ in range(n+1)]

for i in range(m):
    a, b = map(int, input().split())
    graph[a].append(b)
    graph[b].append(a)

# bfs
milestone = [0] * (n+1)
q = []
q.append(1)
while len(q) != 0:
    for i in graph[q[0]]:
        if milestone[i] == 0:
            milestone[i] = q[0]
            q.append(i)
    q.pop(0)

print("Yes")
for i in range(2, n+1):
    print(milestone[i])
Problem:
An elementary school student Takahashi has come to a variety store.  He has two coins, A-yen and B-yen coins (yen is the currency of Japan), and wants to buy a toy that costs C yen. Can he buy it?  Note that he lives in Takahashi Kingdom, and may have coins that do not exist in Japan.

Incorrect submission (verdict: Wrong Answer):
def main(A: int, B: int, C: int):
    if A + B <= C:
        return 'Yes'
    else:
        return 'No'


if __name__ == "__main__":
    A, B, C = map(int, input().split(' '))

    print(main(A, B, C))


Fixed, accepted submission:
def main(A: int, B: int, C: int):
    if (A + B) >= C:
        return 'Yes'
    else:
        return 'No'


if __name__ == "__main__":
    A, B, C = map(int, input().split(' '))

    print(main(A, B, C))
Problem:
Snuke loves working out. He is now exercising N times.  Before he starts exercising, his _power_ is 1. After he exercises for the i-th time, his power gets multiplied by i.  Find Snuke's power after he exercises N times. Since the answer can be extremely large, print the answer modulo 10^{9}+7.

Incorrect submission (verdict: Wrong Answer):
import math
n = int(input())
math.factorial(n)%(10**9 + 7)

Fixed, accepted submission:
ans=1
n=int(input())
for i in range(1,n+1):
    ans*=i
    ans=ans%(10**9+7)
print(ans)
Problem:
There is a game that involves three variables, denoted A, B, and C.  As the game progresses, there will be N events where you are asked to make a choice. Each of these choices is represented by a string s_i. If s_i is `AB`, you must add 1 to A or B then subtract 1 from the other; if s_i is `AC`, you must add 1 to A or C then subtract 1 from the other; if s_i is `BC`, you must add 1 to B or C then subtract 1 from the other.  After each choice, none of A, B, and C should be negative.  Determine whether it is possible to make N choices under this condition. If it is possible, also give one such way to make the choices.

Incorrect submission (verdict: Wrong Answer):
import sys
sys.setrecursionlimit(1000000)

def F(S,curr,D,res):
    
    if len(S)==curr:
        print("Yes")
        for c in res:
            print(c)
        #print("".join(res))
        return True
    s=S[curr]
    
    a=False
    b=False
    if D[s[0]]>=1:
        D[s[0]]-=1
        D[s[1]]+=1
        res.append(s[0])
        a=F(S,curr+1,D,res)
        res.pop(-1)
        D[s[0]]+=1
        D[s[1]]-=1
    else:
        if D[s[1]]>=1:
            D[s[1]]-=1
            D[s[0]]+=1
            res.append(s[1])
            b=F(S,curr+1,D,res)
            res.pop()
            D[s[1]]+=1
            D[s[0]]-=1
    return a or b


N,A,B,C=map(int,input().strip().split())
D={"A":A,"B":B,"C":C}
S=[]
res=[]
for i in range(N):
    S.append(input())
if not (F(S,0,D,res)):
    print("No")

Fixed, accepted submission:
import sys
sys.setrecursionlimit(1000000)

def F(S,curr,D,res):
    if D["A"]<0 or D["B"]<0 or D["C"]<0:
        return False
    if len(S)==curr:
        print("Yes")
        for c in res:
            print(c)
        #print("".join(res))
        return True
    s=S[curr]
    
    a=False
    b=False
    if D[s[0]]>=1:
        D[s[0]]-=1
        D[s[1]]+=1
        res.append(s[1])
        a=F(S,curr+1,D,res)
        res.pop(-1)
        D[s[0]]+=1
        D[s[1]]-=1
    if a:
        return a

    if D[s[1]]>=1:
            D[s[1]]-=1
            D[s[0]]+=1
            res.append(s[0])
            b=F(S,curr+1,D,res)
            res.pop()
            D[s[1]]+=1
            D[s[0]]-=1
    return a or b


N,A,B,C=map(int,input().strip().split())
D={"A":A,"B":B,"C":C}
S=[]
res=[]
for i in range(N):
    S.append(input())
if not (F(S,0,D,res)):
    print("No")