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;
const int mod = 1000000007;
const int LIM = 100005;
long long power(long long x, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) {
      res = res * x % mod;
    }
    x = x * x % mod;
    n >>= 1;
  }
  return (res % mod);
}
int32_t main() {
  long long tt = 1, n, x, k, y, z, i, j;
  cin >> tt;
  while (tt--) {
    string s, w = "", second = "";
    cin >> s;
    cin >> x;
    n = s.length();
    for (i = 0; i < n; i++) {
      w += "1";
      second += "0";
    }
    for (i = 0; i < n; i++) {
      if (s[i] == '0') {
        if (i + x < n) w[i + x] = '0';
        if (i - x >= 0) w[i - x] = '0';
      }
    }
    long long f = 0;
    for (i = 0; i < n; i++) {
      if (w[i] == '1') {
        if (i + x < n) second[i + x] = '1';
        if (i - x >= 0) second[i - x] = '1';
      }
    }
    if (second == s)
      cout << w << '\n';
    else
      cout << -1 << '\n';
  }
}
