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 n = (int)S.size();
  int x;
  cin >> x;
  vector<int> s(n, 0);
  for (int i = 0; i < n; i++) {
    s[i] = S[i] - '0';
  }
  vector<int> res(n, 1);
  bool ok = true;
  for (int i = 0; i < n; i++) {
    int lidx = i - x;
    int ridx = i + x;
    if (s[i] == 0) {
      if (lidx >= 0) res[lidx] = 0;
      if (ridx < n) res[ridx] = 0;
    }
  }
  for (int i = 0; i < n; i++) {
    int lidx = i - x;
    int ridx = i + x;
    int cnt1 = 0;
    if (lidx >= 0 && res[lidx] == 1) cnt1++;
    if (ridx < n && res[ridx] == 1) cnt1++;
    if (s[i]) {
      if (cnt1 == 0) ok = false;
    } else {
      if (cnt1) ok = false;
    }
  }
  if (!ok) {
    cout << -1 << endl;
    return;
  }
  for (int i : res) {
    cout << i;
  }
  cout << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int T;
  T = 1;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}
