Consider the following process. You have a binary string (a string where each character is either 0 or 1) w of length n and an integer x. You build a new binary string s consisting of n characters. The i-th character of s is chosen as follows:

  * if the character w_{i-x} exists and is equal to 1, then s_i is 1 (formally, if i > x and w_{i-x} =  1, then s_i =  1); 
  * if the character w_{i+x} exists and is equal to 1, then s_i is 1 (formally, if i + x ≤ n and w_{i+x} =  1, then s_i =  1); 
  * if both of the aforementioned conditions are false, then s_i is 0. 



You are given the integer x and the resulting string s. Reconstruct the original string w.

Input

The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.

Each test case consists of two lines. The first line contains the resulting string s (2 ≤ |s| ≤ 10^5, each character of s is either 0 or 1). The second line contains one integer x (1 ≤ x ≤ |s| - 1).

The total length of all strings s in the input does not exceed 10^5.

Output

For each test case, print the answer on a separate line as follows:

  * if no string w can produce the string s at the end of the process, print -1; 
  * otherwise, print the binary string w consisting of |s| characters. If there are multiple answers, print any of them. 

Example

Input


3
101110
2
01
1
110
1


Output


111011
10
-1
import sys

def main():
    T = int(sys.stdin.readline())
    while T:
        T -= 1

        A = sys.stdin.readline().rstrip()

        X = int(sys.stdin.readline())

        ans = ['0'] * len(A)

        FLAG = True
        # Numbers that have no reference should be zero
        for i in range(len(A) - X, X):
            if A[i] == '1':
                print(-1)
                FLAG = False
                break

        if not FLAG:
            continue

        for i in range(len(A)):
            if not inBound(A, i - X):
                if not inBound(A, i + X):
                    ans[i] = '0'
                else:
                    ans[i] = A[i + X]
                continue

            if not inBound(A, i + X):
                if not inBound(A, i - X):
                    ans[i] = '0'
                else:
                    ans[i] = A[i - X]
                continue

            if A[i - X] == '1' and (not inBound(A, i - (X << 1)) or ans[i - (X << 1)] == '0'):
                ans[i] = '1'
            elif A[i - X] == '0':
                ans[i] = '0'
            else:
                ans[i] = A[i + X]

            if not verify_backward(A, X, i + X, ans):
                print(-1)
                FLAG = False
                break

        if FLAG:
            print(''.join(ans))

def inBound(A, i):
    return 0 <= i < len(A)

def verify_backward(A, X, i, ans):
    #print(f'Verity. len={len(A)}, A[{i}]={A[i]}, ans[{i-X}]={ans[i-X]}, and[{i+X}]={ans[(i+X)%len(A)]}.')
    if ans[i - X] == '1' and A[i] == '0':
        return False
    return True if inBound(A, i + X) else A[i] == ans[i - X]

if __name__ == '__main__':
    main()
