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 ans[100005];
void solve() {
  string s;
  cin >> s;
  int n = s.length();
  s = "#" + s;
  int x;
  cin >> x;
  memset(ans, -1, sizeof ans);
  for (int i = 1; i <= n; ++i) {
    if (s[i] == '0') {
      if (i - x > 0) ans[i - x] = 0;
      if (i + x <= n) ans[i + x] = 0;
    }
  }
  for (int i = 1; i <= n; ++i)
    if (ans[i] == -1) ans[i] = 1;
  for (int i = 1; i <= n; ++i) {
    bool ok = 1;
    if (s[i] == '0') {
      if (i - x > 0 && ans[i - x] != 0) ok = 0;
      if (i + x <= n && ans[i + x] != 0) ok = 0;
    } else {
      ok = 0;
      if (i - x > 0 && ans[i - x] == 1) ok = 1;
      if (i + x <= n && ans[i + x] == 1) ok = 1;
    }
    if (!ok) {
      cout << -1 << endl;
      return;
    }
  }
  for (int i = 1; i <= n; ++i) cout << ans[i];
  cout << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  while (t--) solve();
}
