TnT/CodeNet4Repair huggingface dataset:
Problem:
Two students of AtCoder Kindergarten are fighting over candy packs.  There are three candy packs, each of which contains a, b, and c candies, respectively.  Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible.  Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.

Incorrect submission (verdict: Wrong Answer):
a,b,c = map(int ,input().split())

if a+b == c :
  print("YES")
elif a+c == b:
  print("YES")
elif b+c == a:
  print("YES")
else :
  print("NO")

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

if a == b+c or b == a+c or c == a+b:
  print("Yes")
else :
  print("No")
Problem:
Two students of AtCoder Kindergarten are fighting over candy packs.  There are three candy packs, each of which contains a, b, and c candies, respectively.  Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible.  Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.

Incorrect submission (verdict: Wrong Answer):
a,b,c=map(int,input().split())
#print(a,b,c)
#print(max(a,b,c))
if max(a,b,c)==(a+b+c)/2:
  print("YES")
else:
  print("NO")

Fixed, accepted submission:
a,b,c=map(int,input().split())
#print(a,b,c)
#print(max(a,b,c))
if max(a,b,c)==(a+b+c)/2:
  print("Yes")
else:
  print("No")
Problem:
You are given a image with a height of H pixels and a width of W pixels. Each pixel is represented by a lowercase English letter. The pixel at the i-th row from the top and j-th column from the left is a_{ij}.  Put a box around this image and output the result. The box should consist of `#` and have a thickness of 1.

Incorrect submission (verdict: Wrong Answer):

mat_small = []
xy = input().split(' ')
for i in range(int(xy[0])):
    mat_small.append(input())

big  = [(int(xy[0])+ 2)* '#'  for a in range(int(xy[0])+ 2) ]

for s in range(1, (int(xy[0])+ 1)):
    big[s] = '#' + mat_small[s-1] + '#'
print( big)          


Fixed, accepted submission:

mat_small = []
xy = input().split(' ')
for i in range(int(xy[0])):
    mat_small.append(input())

big  = [(int(xy[1])+ 2)* '#'  for a in range(int(xy[0])+ 2) ]

for s in range(1, (int(xy[0])+ 1)):
    big[s] = '#' + mat_small[s-1] + '#'

for v in big:
    
    print(v)
Problem:
We say that a odd number N is _similar to 2017_ when both N and (N+1)/2 are prime.  You are given Q queries.  In the i-th query, given two odd numbers l_i and r_i, find the number of odd numbers x similar to 2017 such that l_i ≤ x ≤ r_i.

Incorrect submission (verdict: Time Limit Exceeded):
import numpy as np

def seachPrimeNum(N):
    max = int(np.sqrt(N))
    seachList = [i for i in range(2,N+1)]
    primeNum = []
    while seachList[0] <= max:
        primeNum.append(seachList[0])
        tmp = seachList[0]
        seachList = [i for i in seachList if i % tmp != 0]
    primeNum.extend(seachList)
    return primeNum

#def numsearch(list):
    
        #print(i)
        #if (i+1)/2 not in list:
            #list.remove(i)
    

def main():
    Q = int(input())
    list = seachPrimeNum(100000)
    #list2 = numsearch(list)
    #print(list2)
    c = [0]*100001
    c[0] = 0
    for i in range(1,100001,1):
        #print(i)
        if i in list and (i+1)/2 in list:
            c[i] = c[i-1]+1
        else:
            c[i] = c[i-1]
    #print(c)
    for i in range(Q):
        l,r = map(int,input().split())
        print(c[r]-c[l-1])

main()


Fixed, accepted submission:
import numpy as np

def seachPrimeNum(N):
    max = int(np.sqrt(N))
    seachList = [i for i in range(2,N+1)]
    primeNum = []
    while seachList[0] <= max:
        primeNum.append(seachList[0])
        tmp = seachList[0]
        seachList = [i for i in seachList if i % tmp != 0]
    primeNum.extend(seachList)
    return primeNum

#def numsearch(list):
    
        #print(i)
        #if (i+1)/2 not in list:
            #list.remove(i)
    

def main():
    Q = int(input())
    list = seachPrimeNum(100000)
    
    check = [0]*100001
    primeCheck = [0]*100001
    for i in list:
        primeCheck[i] = 1
    for i in list:
        if i == 2:
            continue
        tmp = int((i+1)/2)
        if primeCheck[tmp] == 1:
            check[i] = 1
    
    c = [0]*100001
    c[0] = 0
    for i in range(1,100001,1):
        #print(i)
        if check[i] == 1:
            c[i] = c[i-1]+1
        else:
            c[i] = c[i-1]
    #print(c)
    for i in range(Q):
        l,r = map(int,input().split())
        print(c[r]-c[l-1])

main()
Problem:
Iroha has a sequence of N strings S_1, S_2, ..., S_N. The length of each string is L.  She will concatenate all of the strings in some order, to produce a long string.  Among all strings that she can produce in this way, find the lexicographically smallest one.  Here, a string s=s_1s_2s_3...s_n is _lexicographically smaller_ than another string t=t_1t_2t_3...t_m if and only if one of the following holds:    * There exists an index i(1≦i≦min(n,m)), such that s_j = t_j for all indices j(1≦j<i), and s_i<t_i.   * s_i = t_i for all integers i(1≦i≦min(n,m)), and n<m.

Incorrect submission (verdict: Wrong Answer):
a = input().split()
a.sort()
print("".join(a))

Fixed, accepted submission:
a=list(map(int,input().split()))
b=[]
for i in range(a[0]):
  b.append(input())
b.sort()
print("".join(b))
Problem:
Takahashi loves numbers divisible by 2.  You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique.  Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder.  For example,    * 6 can be divided by 2 once: 6 -> 3.   * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1.   * 3 can be divided by 2 zero times.

Incorrect submission (verdict: Wrong Answer):
nums = [1, 2, 4, 8, 16, 32, 64]
N = int(input())
result = nums[0]
for num in nums:
    if N > num:
        reuslt = num
    else:
        break
print(result)

Fixed, accepted submission:
nums = [1, 2, 4, 8, 16, 32, 64]
N = int(input())
result = nums[0]
for num in nums:
    if N >= num:
        result = num
    else:
        break
print(result)
Problem:
We have N squares assigned the numbers 1,2,3,\ldots,N. Each square has an integer written on it, and the integer written on Square i is a_i.  How many squares i satisfy both of the following conditions?    * The assigned number, i, is odd.   * The written integer is odd.

Incorrect submission (verdict: Wrong Answer):
N=int(input())
ai=list(map(int,input().split()))

ans=0

for i in range(N):
    if i%2==0:
        pass
    
    if ai[i] %2==1:
        ans+=1
print(ans)


Fixed, accepted submission:
N=int(input())
ai=list(map(int,input().split()))

ans=0

for i in range(N):
    if i%2==1:
        continue

    if ai[i] %2==1:
        ans+=1
print(ans)
Problem:
Two students of AtCoder Kindergarten are fighting over candy packs.  There are three candy packs, each of which contains a, b, and c candies, respectively.  Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible.  Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.

Incorrect submission (verdict: Wrong Answer):
a=list(map(int,input().split()))
print(sum(a))

Fixed, accepted submission:
a,b,c=map(int,input().split())
if a+b==c or a+c==b or b+c==a:
  print("Yes")
else:
  print("No")
Problem:
Write a program which reads a $ n \times m$ matrix $A$ and a $m \times 1$ vector $b$, and prints their product $Ab$.  A column vector with m elements is represented by the following equation.  \\[ b = \left( \begin{array}{c} b_1 \\\ b_2 \\\ : \\\ b_m \\\ \end{array} \right) \\]  A $n \times m$ matrix with $m$ column vectors, each of which consists of $n$ elements, is represented by the following equation.  \\[ A = \left( \begin{array}{cccc} a_{11} & a_{12} & ... & a_{1m} \\\ a_{21} & a_{22} & ... & a_{2m} \\\ : & : & : & : \\\ a_{n1} & a_{n2} & ... & a_{nm} \\\ \end{array} \right) \\]  $i$-th element of a $m \times 1$ column vector $b$ is represented by $b_i$ ($i = 1, 2, ..., m$), and the element in $i$-th row and $j$-th column of a matrix $A$ is represented by $a_{ij}$ ($i = 1, 2, ..., n,$ $j = 1, 2, ..., m$).  The product of a $n \times m$ matrix $A$ and a $m \times 1$ column vector $b$ is a $n \times 1$ column vector $c$, and $c_i$ is obtained by the following formula:  \\[ c_i = \sum_{j=1}^m a_{ij}b_j = a_{i1}b_1 + a_{i2}b_2 + ... + a_{im}b_m \\]

Incorrect submission (verdict: Wrong Answer):
x, y = [int(i) for i in input().split()]
date = [[int(i) for i in input().split()]for i in range(1,x+1)]
date2 =[int(input())for i in range(1,y+1)]
date3 =[]
for k in range(x):
    print(date[k])
print(date2)
for a in range(0,x):
    for b in range(0,y):
        date3 += [date2[b]*date[a][b]]
for h in range(0,len(date3)-1):
    print(date3[h])

Fixed, accepted submission:
x, y = [int(i) for i in input().split()]
date = [[int(i) for i in input().split()]for i in range(1,x+1)]
date2 =[int(input())for i in range(1,y+1)]
date3 =[]
for a in range(0,x):
    for b in range(0,y,):
        date3 += [date2[b]*date[a][b]]
for d in range(0,len(date3),y):
    print(sum(date3[d:d+y]))
Problem:
Print all the integers that satisfies the following in ascending order:    * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.

Incorrect submission (verdict: Wrong Answer):
a,b,k =map(int, input().split())
c=list(range(a,a+k))
d=list(range(b-k+1,b+1))
e=set(c+d)
for i in e:
    if a<=i<=b:
        print(i)

Fixed, accepted submission:
a,b,k =map(int, input().split())
c=list(range(a,a+k))
d=list(range(b-k+1,b+1))
e=list(set(c+d))
e.sort()
for i in e:
    if a<=i<=b:
        print(i)
Problem:
It is November 18 now in Japan. By the way, 11 and 18 are adjacent Lucas numbers.  You are given an integer N. Find the N-th Lucas number.  Here, the i-th Lucas number L_i is defined as follows:    * L_0=2   * L_1=1   * L_i=L_{i-1}+L_{i-2} (i≥2)

Incorrect submission (verdict: Wrong Answer):
n=int(input())
a=2
b=1
c=3
i=0
while i<n:
  d=a
  e=b
  a=b
  b=c
  c=a+b+c
  i=i+1
print(a)

Fixed, accepted submission:
n=int(input())
a=2
b=1
i=0
while i<n:
  d=a
  a=b
  b=d+b
  i=i+1
print(a)
Problem:
Based on some criterion, Snuke divided the integers from 1 through 12 into three groups as shown in the figure below. Given two integers x and y (1 ≤ x < y ≤ 12), determine whether they belong to the same group.  

Incorrect submission (verdict: Wrong Answer):
H,W = [int(i) for i in input().split()]

if H % 3 == 0 or W % 3 == 0:
    print(0)
else:
    s = []
    for i in range(1,H//2 +1):
        j = W//2
        a = [i*W,j*(H-i),(W-j)*(H-i)]
        b = max(a) - min(a)
        s.append(b)
    for i in range(1,W//2+1):
        j = H//2
        a = [H*i,(W-i)*j,(W-i)*(H-j)]
        b = max(a) - min(a)
        s.append(b)

    print(min(s))


Fixed, accepted submission:
A = [1,3,5,7,8,10,12]
B = [4,6,9,11]

a,b = [int(i) for i in input().split()]

if a in A and b in A:
    print('Yes')
elif a in B and b in B:
    print('Yes')
else:
    print('No')
Problem:
Takahashi is competing in a sumo tournament. The tournament lasts for 15 days, during which he performs in one match per day. If he wins 8 or more matches, he can also participate in the next tournament.  The matches for the first k days have finished. You are given the results of Takahashi's matches as a string S consisting of `o` and `x`. If the i-th character in S is `o`, it means that Takahashi won the match on the i-th day; if that character is `x`, it means that Takahashi lost the match on the i-th day.  Print `YES` if there is a possibility that Takahashi can participate in the next tournament, and print `NO` if there is no such possibility.

Incorrect submission (verdict: Wrong Answer):
data = input()
x = data.count("o")
if x + 15-len(data) >= 8:
    print("yes")
else:
    print("no")


Fixed, accepted submission:
data = input()
x = data.count("o")
if x + 15-len(data) >= 8:
    print("YES")
else:
    print("NO")
Problem:
We ask you to select some number of positive integers, and calculate the sum of them.  It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer.  Your objective is to make the sum congruent to C modulo B. Determine whether this is possible.  If the objective is achievable, print `YES`. Otherwise, print `NO`.

Incorrect submission (verdict: Wrong Answer):
import sys
A, B, C = map(int, input().split())
for i in range(1, B - 1):
    if i * A % B == C:
        print('Yes')
        sys.exit()
print('No')

Fixed, accepted submission:
import sys
A, B, C = map(int, input().split())
N = A
while N < A * B:
    if N % B == C:
        print('YES')
        sys.exit()
    N += A
print('NO')
Problem:
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.

Incorrect submission (verdict: Wrong Answer):
import math
a,b=map(int,input().split())
print(math.ceil(a+b/2))

Fixed, accepted submission:
import math
a,b=map(int,input().split())
print(math.ceil((a+b)/2))
Problem:
You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.

Incorrect submission (verdict: Wrong Answer):
s = input()
t = input()
s = ''.join(sorted(s))
t = ''.join(sorted(t, reverse=True))
print(s, t)
print('Yes') if s < t else print('No')


Fixed, accepted submission:
s = input()
t = input()
s = ''.join(sorted(s))
t = ''.join(sorted(t, reverse=True))
print('Yes') if s < t else print('No')
Problem:
You are parking at a parking lot. You can choose from the following two fee plans:    * Plan 1: The fee will be A×T yen (the currency of Japan) when you park for T hours.   * Plan 2: The fee will be B yen, regardless of the duration.  Find the minimum fee when you park for N hours.

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

price = [a * n] + [b]
print(max(price))


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

print(min(a * n, b))
Problem:
Snuke has decided to construct a string that starts with `A` and ends with `Z`, by taking out a substring of a string s (that is, a consecutive part of s).  Find the greatest length of the string Snuke can construct. Here, the test set guarantees that there always exists a substring of s that starts with `A` and ends with `Z`.

Incorrect submission (verdict: Wrong Answer):
#!/usr/bin/env python3
if __name__ == '__main__':
  s = str(input())
  L = len(s)
  beg = -1
  end = -1
  for i in range(0,L):
    if s[i]=='A' and beg==-1:
      beg = i
    elif s[i]=='Z':
      end = i
  print(s[beg:end-beg+1])


Fixed, accepted submission:
#!/usr/bin/env python3
if __name__ == '__main__':
  s = str(input())
  L = len(s)
  beg = -1
  end = -1
  for i in range(0,L):
    if s[i]=='A' and beg==-1:
      beg = i
    elif s[i]=='Z':
      end = i
  print(end-beg+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):
from collections import deque
n, m = map(int, input().split())
ab = [[] for i in range(n+1)]

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

ans = [-1]*(n+1)
que = deque([1])

while que:
    q = que.popleft()
    for i in ab[q]:
        if ans[i] == -1:
            que.append(i)
            ans[i] = q

print("YES")

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

Fixed, accepted submission:
from collections import deque
n, m = map(int, input().split())
ab = [[] for i in range(n+1)]

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

ans = [-1]*(n+1)
que = deque([1])

while que:
  q = que.popleft()
  for i in ab[q]:
    if ans[i] == -1:
      que.append(i)
      ans[i] = q

print("Yes")

for i in range(2,n+1):
  print(ans[i])
Problem:
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order.  To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.

Incorrect submission (verdict: Wrong Answer):
A=list(map(int,input().split()))
print("Yes" if A.count(5)==2 and A.count(7)==1 else "No")

Fixed, accepted submission:
A=list(map(int,input().split()))
print("YES" if A.count(5)==2 and A.count(7)==1 else "NO")
Problem:
Takahashi has N balls. Initially, an integer A_i is written on the i-th ball.  He would like to rewrite the integer on some balls so that there are at most K different integers written on the N balls.  Find the minimum number of balls that Takahashi needs to rewrite the integers on them.

Incorrect submission (verdict: Wrong Answer):
N,K=map(int, input().split())
A=list(map(int, input().split()))
ball={}
result=0
for i in range(N):
  if A[i] in ball.keys():
    ball[A[i]]+=1
  else:
    ball[A[i]]=1
sorted_ball = sorted(list(ball.values()))
if len(sorted_ball) > K:
  for i in range(K):
    result+=1
print(result)

Fixed, accepted submission:
N,K=map(int, input().split())
A=list(map(int, input().split()))
ball={}
result=0
for i in range(N):
  if A[i] in ball.keys():
    ball[A[i]]+=1
  else:
    ball[A[i]]=1
sorted_ball = sorted(list(ball.values()))
if len(sorted_ball) > K:
  for i in range(len(sorted_ball)-K):
    result+=sorted_ball[i]
print(result)
Problem:
Snuke has N sticks. The length of the i-th stick is l_i.  Snuke is making a snake toy by joining K of the sticks together.  The length of the toy is represented by the sum of the individual sticks that compose it. Find the maximum possible length of the toy.

Incorrect submission (verdict: Wrong Answer):
N, K = map(int, input().split())
L = sorted([int(_) for _ in range(N)], reverse=True)

print(sum(L[:K]))

Fixed, accepted submission:
N, K = map(int, input().split())
L = sorted(list(map(int, input().split())), reverse=True)

print(sum(L[:K]))
Problem:
Takahashi is distributing N balls to K persons.  If each person has to receive at least one ball, what is the maximum possible difference in the number of balls received between the person with the most balls and the person with the fewest balls?

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

mi = 1
ma = 1 + k-n

if(k==1):
    print(0)
else:
    print(ma-mi)

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

mi = 1
ma = 1 + n-k

if(k==1):
    print(0)
else:
    print(ma-mi)
Problem:
A Hitachi string is a concatenation of one or more copies of the string `hi`.  For example, `hi` and `hihi` are Hitachi strings, while `ha` and `hii` are not.  Given a string S, determine whether S is a Hitachi string.

Incorrect submission (verdict: Wrong Answer):
hitachi = input(' please input > ')

j = 0
tmp = 0

def check_hit(a,b):
    if a == b:
        return 1
    return 0


for i in hitachi:
    if j % 2 == 0:
        tmp += int(check_hit(i,"h"))
    else:
        tmp += int(check_hit(i,"i"))

if tmp == len(hitachi):
    print("Yse")
else:
    print("No")


Fixed, accepted submission:
hitachi = input()

j = 0
tmp = 0

def check_hit(a,b):
    if a == b:
        return 1
    return 0


for i in hitachi:
    if j % 2 == 0:
        tmp += int(check_hit(i,"h"))
    else:
        tmp += int(check_hit(i,"i"))
    j+=1

if tmp == len(hitachi) and j%2==0 and tmp > 0:
    print("Yes")
else:
    print("No")
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):
x=int(input())
a=int(input())
b=int(input())
print(x-a-b)

Fixed, accepted submission:
x=int(input())
a=int(input())
b=int(input())
v=x-a
c=v//b
print(v-(c*b))
Problem:
E869120 has A 1-yen coins and infinitely many 500-yen coins.   Determine if he can pay exactly N yen using only these coins.

Incorrect submission (verdict: Wrong Answer):
N = int(input())
A = int(input())
if N % 500 <= A:
    print("YES")
else:
    print("NO")

Fixed, accepted submission:
N = int(input())
A = int(input())
if N % 500 <= A:
    print("Yes")
else:
    print("No")
Problem:
There are three houses on a number line: House 1, 2 and 3, with coordinates A, B and C, respectively. Print `Yes` if we pass the coordinate of House 3 on the straight way from House 1 to House 2 without making a detour, and print `No` otherwise.

Incorrect submission (verdict: Wrong Answer):
a, b, c = map(int, input().split())
print('Yes' if (a>c and b<c) or (a<c and b<c) else 'No' )

Fixed, accepted submission:
a, b, c = map(int, input().split())
print('Yes' if (a>c and b<c) or (a<c and b>c) else 'No' )
Problem:
The commonly used bills in Japan are 10000-yen, 5000-yen and 1000-yen bills. Below, the word "bill" refers to only these.  According to Aohashi, he received an otoshidama (New Year money gift) envelope from his grandfather that contained N bills for a total of Y yen, but he may be lying. Determine whether such a situation is possible, and if it is, find a possible set of bills contained in the envelope. Assume that his grandfather is rich enough, and the envelope was large enough.

Incorrect submission (verdict: Wrong Answer):


n, y = map(int, input().split())

#a
#b
#c

i = 0
j = 0
k = 0

l = [-1,-1,-1]
#print(l)

for i in reversed(range(0,n+1)):
    #print('i= %d' % i)
    if  y >= i*10000:
        for j in reversed(range(0,n-i+1)):
            #print('j= %d' % j)
            k = n - i - j
            #print('k= %d' % k)
            if  y == i*10000 + j*5000 + k*1000:
                l = [i, j, k]
                #print(l)

print(l)


Fixed, accepted submission:


n, y = map(int, input().split())

i = 0
j = 0
k = 0

l = [-1,-1,-1]
#print(l)

for i in reversed(range(0,n+1)):
    #print('i= %d' % i)
    if  y >= i*10000:
        for j in reversed(range(0,n-i+1)):
            #print('j= %d' % j)
            k = n - i - j
            #print('k= %d' % k)
            if  y == i*10000 + j*5000 + k*1000:
                l = [i, j, k]


                break
        else:
            continue
        break
#------------------------------------

                #print(l)

print(l[0], l[1], l[2])