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
#!/usr/bin/env python3
# created : 2020. 8. 25. 00:24
import os
from sys import stdin, stdout
 
 
def solve(tc):
    s = stdin.readline().strip()
    x = int(stdin.readline().strip())
    n = len(s)
 
    ans = [1 for i in range(n)]
    for i in range(n):
        if s[i]=='0':
            if i-x>=0:
                ans[i-x] = 0
            if i+x<n:
                ans[i+x] = 0
    for i in range(n):
        if s[i]=='1':
            if (i-x>=0 and ans[i-x]) or (i+x<n and ans[i+x]):
                pass
            else:
                print(-1)
                return
    
    print(*ans, sep='')
 
 
tcs = 1
tcs = int(stdin.readline().strip())
tc = 1
while tc <= tcs:
    solve(tc)
    tc += 1