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, w, t;
  long long n, x, k = 0, cnt = 0, mx = -1000000007, mn = 1000000007;
  cin >> s >> x;
  n = s.length();
  for (int i = 0; i < n; i++) w.push_back('1');
  for (int i = 0; i < n; i++) {
    if (s[i] == '0') {
      if (i >= x) w[i - x] = '0';
      if (i + x < n) w[i + x] = '0';
    }
  }
  bool works = 1;
  for (int i = 0; i < n; i++) {
    if (s[i] == '1') {
      if (i >= x && w[i - x] == '1') continue;
      if (i + x < n && w[i + x] == '1') continue;
      works = 0;
    }
  }
  if (works)
    cout << w;
  else
    cout << -1;
  cout << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long n, tests = 1, prev = 0, i = 0;
  cin >> tests;
  while (tests--) {
    solve();
  }
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  std::mt19937_64 rng(
      std::chrono::steady_clock::now().time_since_epoch().count());
  return 0;
}
