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;
long long mod = 1e9 + 7;
long long min(long long a, long long b) { return (a < b) ? a : b; }
long long max(long long a, long long b) { return (a > b) ? a : b; }
long long fp(long long a, long long b) {
  if (b == 0) return 1;
  long long x = fp(a, b / 2);
  x = (x * x) % mod;
  if (b & 1) x = (x * a) % mod;
  return x;
}
const long long N = 2e5 + 5;
void solve() {
  string s;
  cin >> s;
  long long n = s.size();
  s = '#' + s;
  long long x;
  cin >> x;
  string tmp = "#";
  for (long long i = 1; i <= n; i++) tmp.push_back('1');
  for (long long i = 1; i <= n; i++) {
    if (i + x <= n && s[i] == '0') tmp[i + x] = '0';
    if (i - x >= 1 && s[i] == '0') tmp[i - x] = '0';
  }
  long long f = 1;
  for (long long i = 1; i <= n; i++) {
    long long ok = 0;
    if (i + x <= n && tmp[i + x] == '1') {
      if (s[i] == '0') f = 0;
      ok = 1;
    }
    if (i - x >= 1 && tmp[i - x] == '1') {
      if (s[i] == '0') f = 0;
      ok = 1;
    }
    if (ok == 0) {
      if (s[i] == '1') f = 0;
    }
  }
  if (f) {
    cout << tmp.substr(1) << "\n";
  } else {
    cout << -1 << "\n";
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  srand(time(0));
  long long t;
  cin >> t;
  for (long long i = 1; i <= t; i++) {
    solve();
  }
}
