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;
void solve() {
  string s;
  cin >> s;
  int x;
  cin >> x;
  int n = s.size();
  string p;
  map<int, int> mp;
  for (int i = 0; i < n; i++) p.push_back('%');
  int ans = 0;
  int flag = 0, f1 = 0, f2 = 0;
  for (int i = 0; i < n; i++) {
    flag = 0;
    f1 = 0;
    f2 = 0;
    if (s[i] == '0') {
      if (i - x >= 0) {
        flag = 1;
        if (!mp[i - x]) {
          p[i - x] = '0';
          mp[i - x] = 1;
        } else {
          if (p[i - x] == '1') {
            ans = -1;
            break;
          }
        }
      }
      if (i + x < n) {
        p[i + x] = '0';
        mp[i + x] = 1;
      }
    } else {
      if (i - x >= 0) {
        flag = 1;
        if (!mp[i - x]) {
          p[i - x] = '1';
          mp[i - x] = 1;
        } else if (p[i - x] == '1')
          p[i - x] = '1';
        else {
          f1 = 1;
        }
      }
      if (flag == 0 || f1 == 1) {
        if (i + x < n) {
          mp[i + x] = 1;
          p[i + x] = '1';
        } else {
          f2 = 1;
        }
      }
      if (f2 == 1) {
        ans = -1;
        break;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    if (p[i] == '%') p[i] = '1';
  }
  if (ans == -1)
    cout << "-1" << endl;
  else
    cout << p << endl;
}
int main() {
  int t;
  cin >> t;
  while (t--) solve();
}
