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>
#pragma GCC optimize("Ofast")
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long q;
  cin >> q;
  for (long long i = 0; i < q; i++) {
    string s, w;
    w = "";
    long long n, x;
    cin >> s >> x;
    bool ans = 1;
    n = s.length();
    w.append(s.length(), '1');
    for (long long j = 0; j < n; j++)
      if (s[j] == '1') {
        bool t1 = 0;
        bool t2 = 0;
        if ((j - x) >= 0) {
          if (w[j - x] == '0') t1 = 1;
        }
        if ((j + x) < n) {
          if (w[j + x] == '0') t2 = 1;
        }
        if (t1 && t2) {
          ans = 0;
          break;
        }
      } else {
        if ((j - x) >= 0) w[j - x] = '0';
        if ((j + x) < n) w[j + x] = '0';
      }
    bool t = 1;
    for (long long j = 0; j < n; j++) {
      if (w[j] == '1') {
        if (((j - x) < 0 || s[j - x] == '1') &&
            ((j + x) >= n || s[j + x] == '1')) {
        } else {
          t = 0;
          break;
        }
      }
      if (s[j] == '0') {
        if (((j - x) < 0 || w[j - x] == '0') &&
            ((j + x) >= n || w[j + x] == '0')) {
        } else {
          t = 0;
          break;
        }
      } else {
        if (((j - x) >= 0 && w[j - x] == '1') ||
            ((j + x) < n && w[j + x] == '1')) {
        } else {
          t = 0;
          break;
        }
      }
    }
    if (t && ans)
      cout << w << '\n';
    else
      cout << -1 << '\n';
  }
}
