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>
static const auto fast = []() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(0);
  std::cout.tie(0);
  return 0;
}();
using namespace std;
int32_t main() {
  long long T;
  cin >> T;
  while (T--) {
    string s;
    cin >> s;
    long long k;
    cin >> k;
    long long n = s.size();
    string ans(s.size(), '1');
    for (long long i = 0; i < s.size(); i++) {
      if (s[i] == '0') {
        if (i + k < n) ans[i + k] = '0';
        if (i - k >= 0) ans[i - k] = '0';
      }
    }
    bool flag = 0;
    for (long long i = 0; i < s.size(); i++) {
      if (s[i] == '1') {
        long long z = 0;
        if (i + k >= n || ans[i + k] == '0') {
          z++;
        }
        if (i - k < 0 || ans[i - k] == '0') {
          z++;
        }
        if (z == 2) {
          flag = 1;
          break;
        }
      }
    }
    if (flag) {
      cout << "-1" << endl;
    } else {
      cout << ans << endl;
    }
  }
}
