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 main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    string tmp = "";
    string tmp2 = "";
    int n;
    cin >> n;
    for (int i = 0; i < s.size(); i++) {
      tmp.push_back('1');
      tmp2.push_back('1');
    }
    for (int i = 0; i < s.size(); i++) {
      if (s[i] == '0') {
        int flag = 2;
        int chkpos = i + n;
        if (chkpos >= 0 && chkpos < s.size()) {
          tmp[chkpos] = '0';
        }
        chkpos = i - n;
        if (chkpos >= 0 && chkpos < s.size()) {
          tmp[chkpos] = '0';
        }
      }
    }
    for (int i = 0; i < s.size(); i++) {
      int flag = 2;
      int chkpos = i + n;
      if (chkpos >= 0 && chkpos < s.size()) {
        if (tmp[chkpos] == '1') flag--;
      }
      chkpos = i - n;
      if (chkpos >= 0 && chkpos < s.size()) {
        if (tmp[chkpos] == '1') flag--;
      }
      if (flag == 2) {
        tmp2[i] = '0';
      }
    }
    if (tmp2 == s)
      cout << tmp << '\n';
    else
      cout << -1 << '\n';
  }
}
