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;
int M = 1e9 + 7, N = 205001;
int mm(int x, int y) {
  x %= M, y %= M;
  return (x * y) % M;
}
int po(int x, int y) {
  if (!y) return 1;
  int a = po(x, y / 2) % M;
  if (y % 2) return mm(a, mm(a, x));
  return mm(a, a);
}
void solve() {
  int n, k;
  string s;
  cin >> s >> k;
  n = s.size();
  string a(n, '1');
  for (int i = 0; 0 < n ? i < n : i > n; 0 < n ? i += 1 : i -= 1) {
    if (s[i] == '0' and i + k < n) a[i + k] = '0';
    if (s[i] == '0' and i - k >= 0) a[i - k] = '0';
  }
  for (int i = 0; 0 < n ? i < n : i > n; 0 < n ? i += 1 : i -= 1) {
    if (s[i] == '1')
      if (!((i + k < n and a[i + k] == '1') or
            (i - k >= 0 and a[i - k] == '1'))) {
        cout << "-1\n";
        return;
      }
  }
  cout << a << "\n";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}
