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;
inline long long read() {
  long long nm = 0;
  bool fh = true;
  char cw = getchar();
  for (; !isdigit(cw); cw = getchar()) fh ^= (cw == '-');
  for (; isdigit(cw); cw = getchar()) nm = nm * 10 + (cw - '0');
  return fh ? nm : -nm;
}
char ch[200200], nw[200200];
int n, x, w[200200];
inline void solve() {
  scanf("%s", ch + 1), x = read(), n = strlen(ch + 1);
  for (int i = 1; i <= n; i++) w[i] = 1;
  for (int i = 1; i <= n; i++)
    if (ch[i] == '0') {
      if (i > x) w[i - x] = 0;
      if (i + x <= n) w[i + x] = 0;
    }
  for (int i = 1; i <= n; i++) {
    nw[i] = '0';
    if (i > x && w[i - x] == 1) nw[i] = '1';
    if (i + x <= n && w[i + x] == 1) nw[i] = '1';
  }
  for (int i = 1; i <= n; i++)
    if (nw[i] ^ ch[i]) {
      puts("-1");
      return;
    }
  for (int i = 1; i <= n; i++) printf("%d", w[i]);
  puts("");
}
int main() {
  for (int Cas = read(); Cas; --Cas) solve();
  return 0;
}
