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;
using ll = long long;
template <typename T>
void DEBUG(string label, T value) {
  cerr << "[" << label << " = " << value << "]\n";
}
void solve() {
  string s;
  int x;
  cin >> s >> x;
  int n = (int)s.size();
  string ans(n, ' ');
  for (int i = 0; i < x; ++i) {
    int e = i + x;
    if (e < n) {
      ans[e] = s[i];
    }
  }
  for (int i = n - 1; i >= n - 1 - x + 1; --i) {
    int e = i - x;
    if (e >= 0) {
      if (ans[e] != ' ' && ans[e] != s[i]) {
        cout << -1 << '\n';
        return;
      }
      ans[e] = s[i];
    }
  }
  for (int i = 0; i < n; ++i) {
    if (s[i] == '1') continue;
    int one = i - x;
    int two = i + x;
    if (one < 0 && two >= n) {
      continue;
    }
    if (one >= 0) {
      if (ans[one] == '1') {
        cout << -1 << '\n';
        return;
      } else {
        ans[one] = '0';
      }
    }
    if (two < n) {
      if (ans[two] == '1') {
        cout << -1 << '\n';
        return;
      } else {
        ans[two] = '0';
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    if (s[i] == '0') {
      continue;
    }
    int one = i - x;
    int two = i + x;
    if (one < 0 && two >= n) {
      cout << -1 << '\n';
      return;
    }
    if (one >= 0 && two < n) {
      if (ans[one] == '0' && ans[two] == '0') {
        cout << -1 << '\n';
        return;
      }
    } else if (one >= 0) {
      if (ans[one] == '0') {
        cout << -1 << '\n';
        return;
      }
      ans[one] = '1';
    } else if (two <= n) {
      if (ans[two] == '0') {
        cout << -1 << '\n';
        return;
      }
      ans[two] = '1';
    }
  }
  for (auto i : ans) {
    if (i == ' ') {
      cout << '1';
    } else {
      cout << i;
    }
  }
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}
