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(0);
  cout.tie(0);
  cout.precision(15);
  long long int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    long long int x;
    cin >> x;
    long long int n = (long long int)s.length();
    vector<int> v(n, 2);
    for (long long int i = x; i < n - x; i++) {
      if (s[i - x] == '1' && s[i + x] == '1')
        v[i] = 1;
      else
        v[i] = 0;
    }
    for (long long int i = 0; i < x; i++) {
      if ((i + x) < n && s[i + x] == '0')
        v[i] = 0;
      else
        v[i] = 1;
    }
    for (long long int i = n - x; i < n; i++) {
      if ((i - x) >= 0 && s[i - x] == '0')
        v[i] = 0;
      else
        v[i] = 1;
    }
    string ans;
    for (long long int i = 0; i < n; i++) {
      if (i - x >= 0 && v[i - x] == 1)
        ans += "1";
      else if (i + x < n && v[i + x] == 1)
        ans += "1";
      else
        ans += "0";
    }
    if (ans != s)
      cout << "-1\n";
    else {
      string a;
      for (long long int i = 0; i < n; i++) a += ('0' + v[i]);
      cout << a << "\n";
    }
  }
}
