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;
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
const int INF = 1e9 + 23;
const long long INFL = 2e18 + 23;
inline void solve() {
  int n, x;
  string s;
  cin >> s >> x;
  n = s.size();
  vector<int> ans(n, -1);
  bool ok = true;
  for (int i = 0; i < n; i++) {
    if (s[i] == '1') continue;
    int l = i - x;
    int r = i + x;
    if (l >= 0) ans[l] = 0;
    if (r < n) ans[r] = 0;
  }
  for (int i = 0; i < n; i++) {
    if (s[i] == '0') continue;
    int l = i - x;
    int r = i + x;
    bool done = false;
    if (l >= 0 and (ans[l] == -1 or ans[l] == 1)) ans[l] = 1, done = true;
    if (r < n and (ans[r] == -1 or ans[r] == 1)) ans[r] = 1, done = true;
    if (!done) {
      cout << "-1\n";
      return;
    }
  }
  for (int i = 0; i < n; i++) {
    if (ans[i] == -1) ans[i] = 1;
    cout << ans[i];
  }
  cout << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int tc;
  cin >> tc;
  for (int nc = 1; nc <= tc; nc++) {
    solve();
  }
  return 0;
}
