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
#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
char w[100005], s[100005];
void solve() {
  scanf("%s", s);
  int x;
  scanf("%d", &x);
  int n = strlen(s);
  for (int i = 0; i < n; i++) w[i] = '2';
  for (int i = 0; i < n; i++) {
    if (s[i] == '1') {
      if (i - x >= 0 && w[i - x] == '1' || i + x < n && w[i + x] == '1')
        continue;
      else {
        if (i - x >= 0 && w[i - x] == '2')
          w[i - x] = '1';
        else if (i + x < n && w[i + x] == '2')
          w[i + x] = '1';
        else {
          printf("-1\n");
          return;
        }
      }
    } else {
      if (i - x < 0 && i + x >= n)
        continue;
      else if (i + x >= n && w[i - x] == '0')
        continue;
      else if (i - x < 0 && w[i + x] == '0')
        continue;
      else {
        if (i - x >= 0) {
          if (w[i - x] == '1') {
            puts("-1");
            return;
          } else
            w[i - x] = '0';
        }
        if (i + x < n) {
          if (w[i + x] == '1') {
            puts("-1");
            return;
          } else
            w[i + x] = '0';
        }
      }
    }
  }
  for (int i = 0; i < n; i++) {
    printf("%c", w[i] == '2' ? '0' : w[i]);
  }
  printf("\n");
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) solve();
  return 0;
}
