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 = 1e5 + 5;
int a[MAXN], b[MAXN];
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    getchar();
    int c, x, n = 0;
    while ((c = getchar()) != '\n') a[++n] = c - '0';
    scanf("%d", &x);
    int ok = 1;
    for (int i = 1; i <= n; i++) b[i] = 1;
    for (int i = 1; i <= n; i++) {
      if (!a[i]) {
        if (i > x) b[i - x] = 0;
        if (i + x <= n) b[i + x] = 0;
      }
    }
    for (int i = 1; i <= n; i++) {
      if (a[i]) {
        if (!((i > x && b[i - x]) || (i + x <= n && b[i + x]))) {
          ok = 0;
          break;
        }
      }
    }
    if (ok) {
      for (int i = 1; i <= n; i++) {
        printf("%d", b[i]);
      }
      printf("\n");
    } else
      printf("-1\n");
  }
  return 0;
}
