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;
long long nmod = 998244353;
const double PI = 3.14159265358979323846;
void solve() {
  long long n;
  string s;
  long long x;
  cin >> s >> x;
  n = s.size();
  s = 'G' + s;
  string ans = string(n + 1, 'G');
  for (long long j = 1; j <= n; j++) {
    if (s[j] == '0') {
      if (j - x >= 1) ans[j - x] = '0';
      if (j + x <= n) ans[j + x] = '0';
    } else {
      if (j - x >= 1 && ans[j - x] == 'G') ans[j - x] = '1';
      if (j + x <= n && ans[j + x] == 'G') ans[j + x] = '1';
    }
  }
  for (long long i = 1; i <= n; i++) {
    if (s[i] == '0') {
      if (i - x >= 1 && ans[i - x] == '1') {
        cout << -1;
        return;
      } else if (i + x <= n && ans[i + x] == '1') {
        cout << -1;
        return;
      }
    } else {
      if ((i - x >= 1 && ans[i - x] == '1') ||
          (i + x <= n && ans[i + x] == '1')) {
        continue;
      } else {
        cout << -1;
        return;
      }
    }
  }
  string res = ans.substr(1, n);
  for (long long i = 0; i < n; i++)
    if (res[i] == 'G') res[i] = '0';
  cout << res;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t;
  cin >> t;
  while (t--) {
    solve();
    cout << endl;
  }
  return 0;
}
