TnT/CodeNet4Repair huggingface dataset:
Problem:
You are given an H × W grid.   The squares in the grid are described by H strings, S_1,...,S_H.   The j-th character in the string S_i corresponds to the square at the i-th row from the top and j-th column from the left (1 \leq i \leq H,1 \leq j \leq W).   `.` stands for an empty square, and `#` stands for a square containing a bomb.  Dolphin is interested in how many bomb squares are horizontally, vertically or diagonally adjacent to each empty square.   (Below, we will simply say "adjacent" for this meaning. For each square, there are at most eight adjacent squares.)   He decides to replace each `.` in our H strings with a digit that represents the number of bomb squares adjacent to the corresponding empty square.  Print the strings after the process.

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

field = []

for i in range(h):
	field.append( input())

for i in range(h):
	for j in range(w):
		if field[i][j] == ".":
			hmin = max(i-1, 0)
			hmax = min(i+1, h-1)
			wmin = max(j-1, 0)
			wmax = min(j+1, w-1)
			count = 0
			for m in range(hmin, hmax+1):
				for n in range(wmin, wmax+1):
					if field[m][n] == "#":
						count+=1

			print (count)
			# field[i][j] = count
			field[i] = field[i][0:j]+ str(count) + field[i][j+1:]


for i in range(h):
	print (field[i])


Fixed, accepted submission:
h, w=[int(i) for i in input().split()]

field = []

for i in range(h):
	field.append( input())

for i in range(h):
	for j in range(w):
		if field[i][j] == ".":
			hmin = max(i-1, 0)
			hmax = min(i+1, h-1)
			wmin = max(j-1, 0)
			wmax = min(j+1, w-1)
			count = 0
			for m in range(hmin, hmax+1):
				for n in range(wmin, wmax+1):
					if field[m][n] == "#":
						count+=1

#			print (count)
			# field[i][j] = count
			field[i] = field[i][0:j]+ str(count) + field[i][j+1:]


for i in range(h):
	print (field[i])
Problem:
You are given an undirected connected graph with N vertices and M edges that does not contain self-loops and double edges.   The i-th edge (1 \leq i \leq M) connects Vertex a_i and Vertex b_i.  An edge whose removal disconnects the graph is called a _bridge_.   Find the number of the edges that are bridges among the M edges.

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

road = [[] for _ in range(N)]

for _ in range(M):
  a,b = map(int,input().split())
  road[a-1].append(b-1)
  road[b-1].append(a-1)

def dfs(v,w):
  l = [0 for _ in range(N)]
  l[v] = 1
  stack = [v]
  while stack:
    vv = stack.pop()
    for i in road[vv]:
      if i == w and v == vv or i == vv and v == w:
        continue
      else:
        if l[i] == 0:
          l[i] = 1
          stack.append(i)
  return sum(l)

ans = 0
check = [[0 for _ in range(N)] for _ in range(N)]

for n in range(N):
  for m in range(len(road[n])):
    if check[n][road[n][m]] == 0 or check[road[n][m]][n] == 0:
      check[n][road[n][m]] = 1
      check[road[n][m]][n] = 1
      d = dfs(n,road[n][m])
      if d != N:
        ans += 1

print(ans,road)

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

road = [[] for _ in range(N)]

for _ in range(M):
  a,b = map(int,input().split())
  road[a-1].append(b-1)
  road[b-1].append(a-1)

def dfs(v,w):
  l = [0 for _ in range(N)]
  l[v] = 1
  stack = [v]
  while stack:
    vv = stack.pop()
    for i in road[vv]:
      if i == w and v == vv or i == vv and v == w:
        continue
      else:
        if l[i] == 0:
          l[i] = 1
          stack.append(i)
  return sum(l)

ans = 0
check = [[0 for _ in range(N)] for _ in range(N)]

for n in range(N):
  for m in range(len(road[n])):
    if check[n][road[n][m]] == 0 or check[road[n][m]][n] == 0:
      check[n][road[n][m]] = 1
      check[road[n][m]][n] = 1
      d = dfs(n,road[n][m])
      if d != N:
        ans += 1

print(ans)
Problem:
Two deer, AtCoDeer and TopCoDeer, are playing a game called _Honest or Dishonest_. In this game, an honest player always tells the truth, and an dishonest player always tell lies. You are given two characters a and b as the input. Each of them is either `H` or `D`, and carries the following information:  If a=`H`, AtCoDeer is honest; if a=`D`, AtCoDeer is dishonest. If b=`H`, AtCoDeer is saying that TopCoDeer is honest; if b=`D`, AtCoDeer is saying that TopCoDeer is dishonest.  Given this information, determine whether TopCoDeer is honest.

Incorrect submission (verdict: Time Limit Exceeded):
X = map(int,input().split())
#X=7
P = {0}
i=1
while True:
    Q = {0}
    for p in P:
        Q.add(p+i)
        Q.add(p-i)
    P = P.union(Q)
    if X in P:
        print(i)
        break
    i = i+1


Fixed, accepted submission:
a,b = input().split()
if a == b:
     print('H')
else:
     print('D')
Problem:
AtCoDeer the deer recently bought three paint cans. The color of the one he bought two days ago is a, the color of the one he bought yesterday is b, and the color of the one he bought today is c. Here, the color of each paint can is represented by an integer between 1 and 100, inclusive.  Since he is forgetful, he might have bought more than one paint can in the same color. Count the number of different kinds of colors of these paint cans and tell him.

Incorrect submission (verdict: Wrong Answer):
count = 0

a, b, c = map(int, input().split())
if a == b:
    count += 1
if b == c:
    count += 1
if b == c:
    count += 1
    
print(count)


Fixed, accepted submission:
count = 3
a, b, c = map(int, input().split())
if a == b:
    count -= 1
if a == c:
    count -= 1
if b == c:
    count -= 1
if count == 0:
    count += 1
print(count)
Problem:
When l is an odd number, the median of l numbers a_1, a_2, ..., a_l is the (\frac{l+1}{2})-th largest value among a_1, a_2, ..., a_l.  You are given N numbers X_1, X_2, ..., X_N, where N is an even number. For each i = 1, 2, ..., N, let the median of X_1, X_2, ..., X_N excluding X_i, that is, the median of X_1, X_2, ..., X_{i-1}, X_{i+1}, ..., X_N be B_i.  Find B_i for each i = 1, 2, ..., N.

Incorrect submission (verdict: Wrong Answer):
n = int(input())
x = map(int, input().split())
temp = sorted(x)
idx = n//2
y = temp[idx-1]
z = temp[idx]
for item in x:
    if item < y:
        print(y)
    else:
        print(z)

Fixed, accepted submission:
n = int(input())
x = list(map(int, input().split()))
temp = sorted(x)
idx = n//2
y = temp[idx-1]
z = temp[idx]
for item in x:
    if item >= z:
        print(y)
    else:
        print(z)
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):
h,w = map(int,input().split())
l=[str(input()) for i in range(h)]
p="*"
print(p*(w+2))
for j in l:
    print(p+j+p)
print(p*(w+2))

Fixed, accepted submission:
h,w = map(int,input().split())
l=[str(input()) for i in range(h)]
p="#"
print(p*(w+2))
for j in l:
    print(p+j+p)
print(p*(w+2))
Problem:
We have a 3×3 square grid, where each square contains a lowercase English letters. The letter in the square at the i-th row from the top and j-th column from the left is c_{ij}.  Print the string of length 3 that can be obtained by concatenating the letters in the squares on the diagonal connecting the top-left and bottom-right corner of the grid, from the top-left to bottom-right.

Incorrect submission (verdict: Wrong Answer):
a = input()
b = input()
c = input()

A = [0]*3
ca = 0
cb = 0
cc = 0
for d in a:
    ca +=1
    if(ca == 1):
        A[0] = d

for e in b:
    cb +=1
    if(cb == 1):
        A[1] = e
for f in c:
    cc +=1
    if(cc == 1):
        A[2] = f 

print(A[0]+A[1]+A[2])

Fixed, accepted submission:
a = input()
b = input()
c = input()
print(a[0] + b[1] + c[2])
Problem:
Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions.  You are given two strings O and E. O contains the characters at the odd- numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password.

Incorrect submission (verdict: Wrong Answer):
o, e = [list(input()) for _ in range(2)]
ans = []
for i in range(len(o)):
  ans.append(o[i])
  if len(o)==len(e):
    ans.append(e[i])
for j in ans:
  print(j, end='')

Fixed, accepted submission:
o, e = [list(input()) for _ in range(2)]
ans = ''
for i in range(len(o)):
  ans += o[i]
  if i == len(o)-1 and len(o) != len(e):
    break
  ans += e[i]
print(ans)
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):
a, b, c = map(int, input().split())

for i in range(a, a * b + 1, a):
    if i % b == c:
        print('Yes')
        break
else:
    print('No')

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

for i in range(1, 1001):
    n = i * a
    if n % b == c:
        print('YES')
        break
else:
    print('NO')
Problem:
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.

Incorrect submission (verdict: Wrong Answer):
a,b = map(int,input().split())
c = a*b % 2
if c == 0:
    print("even")
else:
    print("odd")


Fixed, accepted submission:
a,b = map(int,input().split())
c = a*b % 2
if c == 0:
    print("Even")
else:
    print("Odd")
Problem:
Given is a rooted tree with N vertices numbered 1 to N. The root is Vertex 1, and the i-th edge (1 \leq i \leq N - 1) connects Vertex a_i and b_i.  Each of the vertices has a counter installed. Initially, the counters on all the vertices have the value 0.  Now, the following Q operations will be performed:    * Operation j (1 \leq j \leq Q): Increment by x_j the counter on every vertex contained in the subtree rooted at Vertex p_j.  Find the value of the counter on each vertex after all operations.

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


ki = [0] * (n + 1)
for i in range(n - 1):
    a, b = map(int, input().split())
    ki[b] = a
# print("ki=",ki)


c = [0] * (n + 1)


for i in range(q):
    p, x = map(int, input().split())
    c[p] += x
# print("c=",c)

の点数を足す
for i in range(1, n + 1):
    c[i] += c[ki[i]]
# print(c)

print(" ".join(map(str, c[1:])))

Fixed, accepted submission:
import sys
sys.setrecursionlimit(10**7)
input = sys.stdin.readline

N, Q = map(int,input().split())

# AB = [[int(x) for x in input().split()] for _ in range(N-1)]




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


for _ in range(N-1):
    a, b = map(int, input().split())
    graph[a].append(b)
    graph[b].append(a)

# print(graph)
counter = [0] * (N+1)

# print(counter)

# PX = [[int(x) for x in input().split()] for _ in range(Q)]


for _ in range(Q):
  p, x = map(int, input().split())
  counter[p] += x


def shori(pa,c,val):
  val += counter[c]
  counter[c] = val
  for i in graph[c]:
    if i == pa:
      continue
    shori(c,i,val)
  
shori(0,1,0)
  

print(" ".join(map(str, counter[1:])))
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())

l0 = 2
l1 = 1
if n == 1:
    print(l0)
elif n == 2:
    print(l1)
else:
    ans = 0
    for i in range(n-2):
        ans = l0 + l1
        l0 = l1
        l1 = ans
    print(ans)


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

l0 = 2
l1 = 1
if n == 1:
    print(l1)
else:
    ans = 0
    for i in range(n-1):
        ans = l0 + l1
        l0 = l1
        l1 = ans
    print(ans)
Problem:
In this problem, we use the 24-hour clock.  Takahashi gets up exactly at the time H_1 : M_1 and goes to bed exactly at the time H_2 : M_2. (See Sample Inputs below for clarity.) He has decided to study for K consecutive minutes while he is up. What is the length of the period in which he can start studying?

Incorrect submission (verdict: Wrong Answer):
print(-1)

Fixed, accepted submission:

def main():
    line = list(map(int,input().split()))
    H1 = line[0]
    M1 = line[1]
    H2 = line[2]
    M2 = line[3]
    K = line[4]
    h = H2-H1
    m = M2 - M1
    if m < 0:
        h -= 1
        m %= 60
    k = h*60+m-K
    print(k)


if __name__ == "__main__":
    main()
Problem:
We have N sticks with negligible thickness. The length of the i-th stick is A_i.  Snuke wants to select four different sticks from these sticks and form a rectangle (including a square), using the sticks as its sides. Find the maximum possible area of the rectangle.

Incorrect submission (verdict: Wrong Answer):

N=int(input()) 
A=list(map(int,input().split()))
A.sort 
max1=0
max2=0
max3=0
max4=0
for i in range(0,len(A)): 
    if A.count(A[i])>=2: 
        a=max1
        b=max2
        c=max3
        max1=A[i]
        max2=a
        max3=b
        max4=c
print(max1*max4)

Fixed, accepted submission:

N=int(input()) 
A=list(map(int,input().split()))
A.sort() 
max1=0
max2=0
if A[0]==A[1]:
        a=max1
        max1=A[0]
        max2=a     
for i in range(2,len(A)-1): 
    if A[i]==A[i+1] and (A[i]!=A[i-1] or  A[i]==A[i-1]==A[i-2]):
        a=max1
        max1=A[i]
        max2=a 
print(max1*max2)
Problem:
Write a program of the Bubble Sort algorithm which sorts a sequence _A_ in ascending order. The algorithm should be based on the following pseudocode:                 BubbleSort(A) 1 for i = 0 to A.length-1 2     for j = A.length-1 downto i+1 3         if A[j] < A[j-1] 4             swap A[j] and A[j-1]   Note that, indices for array elements are based on 0-origin.  Your program should also print the number of swap operations defined in line 4 of the pseudocode.

Incorrect submission (verdict: Wrong Answer):
def sort(A, exchange):
    for i in range(0, len(A) - 1):
        if modify_order(A, i) == 0:
            break

def modify_order(A, i):
    exchange = 0
    for j in  range(len(A) - 1, i, -1):
        if A[j - 1] > A[j]:
            A[j - 1], A[j] = A[j], A[j - 1]
            exchange = exchange + 1

n = int(input())
s = list(map(int, input().strip().split(' ')))
e = 0
sort(s, e)
print(s)
print(e)



Fixed, accepted submission:
n = int(input())
s = list(map(int, input().strip().split(' ')))
e = int()
e = 0
flag = True
while flag:
        flag = False
        for i in range(n - 1, 0, -1):
            if s[i] < s[i - 1]:
                s[i - 1], s[i] = s[i], s[i - 1]
                e = e + 1
                flag = True

for i in range(0 , len(s)):
    if i != len(s) - 1:
        print(s[i], end=" ")
    else:
        print(s[i])

print(e)
Problem:
Let {\rm comb}(n,r) be the number of ways to choose r objects from among n objects, disregarding order. From n non-negative integers a_1, a_2, ..., a_n, select two numbers a_i > a_j so that {\rm comb}(a_i,a_j) is maximized. If there are multiple pairs that maximize the value, any of them is accepted.

Incorrect submission (verdict: Wrong Answer):
from math import factorial
n=int(input())
a=list(map(int,input().split()))
def comb(n,r):
    return factorial(n)//factorial(n-r)//factorial(r)
print(max([comb(max(a),i) for i in a]))

Fixed, accepted submission:
n=int(input())
a=list(map(int,input().split()))
n=max(a)
a.remove(n)
t=n
for i in a:
    if(t>abs(n/2-i)):
        t=abs(n/2-i)
        r=i
print(n,r)
Problem:
We have an H \times W grid whose squares are painted black or white. The square at the i-th row from the top and the j-th column from the left is denoted as (i, j).   Snuke would like to play the following game on this grid. At the beginning of the game, there is a character called Kenus at square (1, 1). The player repeatedly moves Kenus up, down, left or right by one square. The game is completed when Kenus reaches square (H, W) passing only white squares.   Before Snuke starts the game, he can change the color of some of the white squares to black. However, he cannot change the color of square (1, 1) and (H, W). Also, changes of color must all be carried out before the beginning of the game.   When the game is completed, Snuke's score will be the number of times he changed the color of a square before the beginning of the game. Find the maximum possible score that Snuke can achieve, or print -1 if the game cannot be completed, that is, Kenus can never reach square (H, W) regardless of how Snuke changes the color of the squares.  The color of the squares are given to you as characters s_{i, j}. If square (i, j) is initially painted by white, s_{i, j} is `.`; if square (i, j) is initially painted by black, s_{i, j} is `#`.

Incorrect submission (verdict: Wrong Answer):
H,W = map(int, input().split())
S = [input() for i in range(H)]
queue = [(0,0,1)]
done ={(0,0)}
count = sum(row.count("#") for row in S)

def bfs(x,y,c):
    if(x<0 or y<0 or x>-W or y>=H or S[y][x]=="#"):
        return
    
    if(x,y) in done:
        return
    
    queue.append((x,y,c+1))
    done.add((x,y))

dx = [1,-1,0,0]
dy = [0,0,1,-1] 
flag= -1
while queue:
    x,y,c = queue.pop(0)
    if x==W-1 and y==H-1:
        flag=1
        break
    for i in range(4):
        bfs(x+dx[i],y+dy[i],c)

print(flag if flag == -1 else (H*W-count-c))

Fixed, accepted submission:
H,W = map(int, input().split())
S = [input() for i in range(H)]
queue = [(0,0,1)]
done ={(0,0)}
count = sum(row.count("#") for row in S)

def bfs(x,y,c):
    if(x<0 or y<0 or x>=W or y>=H or S[y][x]=="#"):
        return
    
    if(x,y) in done:
        return
    
    queue.append((x,y,c+1))
    done.add((x,y))

dx = [1,-1,0,0]
dy = [0,0,1,-1] 
flag= -1
while queue:
    x,y,c = queue.pop(0)
    if x==W-1 and y==H-1:
        flag=1
        break
    for i in range(4):
        bfs(x+dx[i],y+dy[i],c)

print(flag if flag == -1 else (H*W-count-c))
Problem:
Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal **integers** by transforming some of them.  He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2).  Find the minimum total cost to achieve his objective.

Incorrect submission (verdict: Wrong Answer):
n = int(input())
a = list(map(int,input().split()))
mn = min(a)
mx = max(a)+1
ans = 0
for i in range(mn, mx):
  cost = 0
  for aa in a:
    cost += (i - aa)**2
  if ans > cost:
    ans = cost
print(ans)

Fixed, accepted submission:
n = int(input())
a = list(map(int,input().split()))
mn = min(a)
mx = max(a)+1
ans = 10**9
for i in range(mn, mx):
  cost = 0
  for aa in a:
    cost += (i - aa)**2
  if ans > cost:
    ans = cost
print(ans)
Problem:
Given is a directed graph G with N vertices and M edges.   The vertices are numbered 1 to N, and the i-th edge is directed from Vertex A_i to Vertex B_i.   It is guaranteed that the graph contains no self-loops or multiple edges.  Determine whether there exists an induced subgraph (see Notes) of G such that the in-degree and out-degree of every vertex are both 1. If the answer is yes, show one such subgraph.   Here the null graph is not considered as a subgraph.

Incorrect submission (verdict: Wrong Answer):

from collections import deque
def resolve():
    N, M = map(int, input().split())
    G = [[] for _ in range(N)]
    for _ in range(M):
        a, b = map(lambda x: int(x) - 1, input().split())
        G[a].append(b)

    shortest = N + 1
    res = []
    for s in range(N):
        dist = [-1] * N
        pre = [-1] * N
        q = deque()
        q.append(s)
        dist[s] = 0
        while q:
            v = q.popleft()
            for to in G[v]:
                if dist[to] == -1:
                    dist[to] = dist[v] + 1
                    pre[to] = v
                    q.append(to)

        for t in range(N):
            if t == s or dist[t] == -1:
                continue
            for to in G[t]:
                if to == s:  
                    tmp = [s]
                    cur = t
                    while cur != s:
                        tmp.append(cur)
                        cur = pre[cur]
                        if shortest > len(tmp):
                            shortest = len(tmp)
                            res = tmp

    if shortest == N + 1:
        print(-1)
    else:
        print(len(res))
        for v in res:
            print(v + 1)


if __name__ == "__main__":
    resolve()


Fixed, accepted submission:

from collections import deque
def resolve():
    N, M = map(int, input().split())
    G = [[] for _ in range(N)]
    for _ in range(M):
        a, b = map(lambda x: int(x) - 1, input().split())
        G[a].append(b)

    shortest = N + 1
    res = []
    for s in range(N):
        dist = [-1] * N
        pre = [-1] * N
        q = deque()
        q.append(s)
        dist[s] = 0
        while q:
            v = q.popleft()
            for to in G[v]:
                if dist[to] == -1:
                    dist[to] = dist[v] + 1
                    pre[to] = v
                    q.append(to)

        for t in range(N):
            if t == s or dist[t] == -1:
                continue
            for to in G[t]:
                if to == s:  
                    tmp = [s]
                    cur = t
                    while cur != s:
                        tmp.append(cur)
                        cur = pre[cur]
                    if shortest > len(tmp):
                        shortest = len(tmp)
                        res = tmp

    if shortest == N + 1:
        print(-1)
    else:
        print(len(res))
        res.sort()
        for v in res:
            print(v + 1)


if __name__ == "__main__":
    resolve()
Problem:
There is a hotel with the following accommodation fee:    * X yen (the currency of Japan) per night, for the first K nights   * Y yen per night, for the (K+1)-th and subsequent nights  Tak is staying at this hotel for N consecutive nights. Find his total accommodation fee.

Incorrect submission (verdict: Wrong Answer):
from collections import Counter

w=list(str(input()))
h=set(w)

for i in h:
   if  w.count(i)%2!=0:
        print("No")
        exit()

print("Yes")


Fixed, accepted submission:
# -*- coding utf-8 -*-

n = int(input())
k = int(input())
x = int(input())
y = int(input())

if n <= k :
    print(n*x)
elif k < n :
    ans = (k*x) + (y*(n-k))
    print(ans)
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):
l=(int,input().split())
f=len([i for i in l if i == 5])==2
s=len([i for i in l if i == 7])==1
print( 'YES' if f and s else 'NO')

Fixed, accepted submission:
def k(a,i1,i2):
  if(a==7):
    i1 = i1+1
  if(a==5):
    i2 = i2+1
  return i1,i2
  
a,b,c=map(int,input().split())
i1=0
i2=0
i1,i2 = k(a,i1,i2)
i1,i2 = k(b,i1,i2)
i1,i2 = k(c,i1,i2)
print('YES' if i1==1 and i2==2 else 'NO')