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 z;
  cin >> z;
  for (int i = 0; i < z; i++) {
    string s;
    cin >> s;
    int x;
    cin >> x;
    int l = s.size();
    string s1(l, '1');
    string s2(l, '1');
    for (int j = 1; j <= l; j++) {
      if (s[j - 1] == '0') {
        if (j - x > 0) {
          s1[j - x - 1] = '0';
        }
        if (j + x <= l) {
          s1[j + x - 1] = '0';
        }
      }
    }
    for (int k = 1; k <= l; k++) {
      if (k - x > 0 && s1[k - x - 1] == '1') {
        continue;
      } else if (k + x <= l && s1[k + x - 1] == '1') {
        continue;
      } else {
        s2[k - 1] = '0';
      }
    }
    int nan = 1;
    for (int p = 0; p < l; p++) {
      if (s2[p] != s[p]) {
        nan = 0;
        break;
      }
    }
    if (nan == 1) {
      cout << s1 << endl;
    } else {
      cout << -1 << endl;
    }
  }
  return 0;
}
