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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t, x;
  cin >> t;
  for (int i = 0; i < t; i++) {
    string s;
    cin >> s >> x;
    bool possible = true;
    for (int j = 0; j < s.size(); j++) {
      if (s[j] == '1') {
        possible = false;
        if (j > x - 1) {
          if (j < 2 * x or s[j - 2 * x] == '1') possible = true;
        }
        if (j < s.size() - x) {
          if (j > (int)s.size() - 2 * x - 1 or s[j + 2 * x] == '1')
            possible = true;
        }
        if (!possible) {
          break;
        }
      }
    }
    if (!possible) {
      cout << -1 << "\n";
    } else {
      for (int j = 0; j < s.size(); j++) {
        if ((j > x - 1 and s[j - x] == '0') or
            (j < s.size() - x and s[j + x] == '0')) {
          cout << 0;
        } else {
          cout << 1;
        }
      }
      cout << "\n";
    }
  }
}
