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;
char s[100005], w[100005];
bool vis[100005];
int main() {
  int ts;
  scanf("%d", &ts);
  while (ts--) {
    int n, x;
    scanf("%s %d", s, &x);
    n = strlen(s);
    for (int i = 0; i < n; i++) vis[i] = 0;
    bool f = 1;
    for (int i = 0; i < n && f; i++) {
      if (s[i] == '1') {
        if ((i - x) >= 0 || (i + x) < n) {
          if ((i - x) >= 0 && (!vis[i - x] || w[i - x] == '1'))
            vis[i - x] = 1, w[i - x] = '1';
          else if ((i + x) < n && (!vis[i + x] || w[i + x] == '1'))
            vis[i + x] = 1, w[i + x] = '1';
          else
            f = 0;
        } else
          f = 0;
      } else {
        if ((i - x) >= 0 && vis[i - x] && w[i - x] == '1')
          f = 0;
        else if ((i + x) < n && vis[i + x] && w[i + x] == '1')
          f = 0;
        else {
          if ((i - x) >= 0) vis[i - x] = 1, w[i - x] = '0';
          if ((i + x) < n) vis[i + x] = 1, w[i + x] = '0';
        }
      }
    }
    for (int i = 0; i < n; i++)
      if (!vis[i]) w[i] = '0';
    if (f) {
      for (int i = 0; i < n; i++) printf("%c", w[i]);
      puts("");
    } else
      puts("-1");
  }
  return 0;
}
