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;
using ll = long long int;
ll a[200004], b[200005];
ll fnc(ll n) {
  ll cnt;
  return cnt;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    ll n, m, sum = 0, cnt = 0;
    string s, s1;
    map<ll, ll> mp;
    cin >> s >> n;
    for (int i = 0; i < s.size(); i++) {
      if (s[i] == '0') {
        ll x = i + 1;
        x += n;
        mp[x] = 1;
        if (x - 2 * n > 0) mp[x - 2 * n] = 1;
      }
    }
    for (int i = 0; i < s.size(); i++) {
      if (s[i] == '1') {
        ll x = i + 1;
        if ((x + n <= s.size() and mp[x + n] == 0 ||
             x - n > 0 and mp[x - n] == 0)) {
        } else {
          cnt = -1;
          break;
        }
      }
    }
    if (cnt == -1)
      cout << -1 << endl;
    else {
      ll x = s.size();
      for (int i = 1; i <= x; i++) {
        if (mp[i] == 1)
          cout << 0;
        else
          cout << 1;
      }
      cout << endl;
    }
  }
}
