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 long long mod = 1e9 + 7;
const long long modx = 998244353;
const long long Nx = 1e5 + 2;
const long long inf = (long long)(1e19 + 10LL);
void solve() {
  string str;
  cin >> str;
  long long x;
  cin >> x;
  long long n = (long long)(str.size());
  vector<char> ans(n, '1');
  for (long long i = 0; i < (long long)(str.size()); i++) {
    if (str[i] == '0') {
      if (i - x >= 0) {
        ans[i - x] = '0';
      }
      if (i + x < n) {
        ans[i + x] = '0';
      }
    }
  }
  string dup = "";
  for (long long i = 0; i < (long long)(ans.size()); i++) {
    if ((i - x >= 0 && ans[i - x] == '1') || (i + x < n && ans[i + x] == '1'))
      dup = dup + "1";
    else
      dup = dup + "0";
  }
  if (str != dup) {
    cout << "-1" << endl;
    return;
  }
  for (auto it : ans) cout << it;
  cout << endl;
  return;
}
signed main() {
  long long test = 1;
  cin >> test;
  while (test--) {
    solve();
  }
  return 0;
}
