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 IMPLEMENT_CODE() {
  char s[200005];
  cin >> s;
  long long int l = strlen(s);
  long long int x;
  cin >> x;
  char ans[l + 1];
  char o = '1';
  char z = '0';
  for (long long int i = 0; i < l; ++i) {
    ans[i] = z;
  }
  ans[l] = '\0';
  for (long long int i = 0; i < l; i++) {
    if (s[i] == '1') {
      if (i + x < l) {
        if (i + 2 * x < l && s[i + 2 * x] == '1')
          ans[i + x] = o;
        else if (i + 2 * x >= l)
          ans[i + x] = o;
      }
      if (i - x >= 0) {
        if (i - 2 * x >= 0 && s[i - 2 * x] == '1')
          ans[i - x] = o;
        else if (i - 2 * x < 0)
          ans[i - x] = o;
      }
    }
  }
  char p[l + 1];
  p[l] = '\0';
  for (long long int i = 0; i < l; ++i) {
    p[i] = z;
  }
  for (long long int i = 0; i < l; i++) {
    if (i + x < l && ans[i + x] == '1') p[i] = o;
    if (i - x >= 0 && ans[i - x] == '1') p[i] = o;
  }
  if (strcmp(p, s) == 0)
    cout << ans << endl;
  else {
    int q = 1;
    for (long long int i = 0; i < l; i++) {
      if (s[i] == o && p[i] != o) {
        q = 0;
        break;
      }
    }
    if (q == 1)
      cout << ans << endl;
    else
      cout << -1 << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(0);
  ;
  long long int testCases;
  cin >> testCases;
  while (testCases--) {
    IMPLEMENT_CODE();
  }
  return 0;
}
