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;
const int MOD = (int)1e9 + 7;
const int inf = (int)1e9;
const long long linf = (long long)1e18 + 1;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    int x;
    cin >> x;
    string t(s.length(), '1');
    int n = s.length();
    for (int i = 0; i < n; i++) {
      if (s[i] == '0') {
        if (i + x < n) t[i + x] = '0';
        if (i - x >= 0) t[i - x] = '0';
      }
    }
    bool ok = true;
    for (int i = 0; i < n; i++) {
      if (s[i] == '1') {
        bool f1 = 1, f2 = 1;
        if (((i + x) >= n) || (t[i + x] == '0')) f1 = 0;
        if (((i - x) < 0) || (t[i - x] == '0')) f2 = 0;
        if (!f1 && !f2) ok = false;
      }
    }
    if (!ok)
      cout << -1 << endl;
    else
      cout << t << endl;
  }
  return 0;
}
