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>
template <class T>
void rd(T &x) {
  x = 0;
  T f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
  x *= f;
}
using namespace std;
const int N = 1e5 + 10;
int T;
char s[N];
int x, n;
int ans[N];
int main() {
  rd(T);
  while (T--) {
    scanf("%s", s + 1);
    n = strlen(s + 1);
    rd(x);
    for (int i = 1; i <= n; i++) ans[i] = 1;
    for (int i = 1; i <= n; i++) {
      if (s[i] - '0') continue;
      if (i > x) ans[i - x] = 0;
      if (i + x <= n) ans[i + x] = 0;
    }
    bool flg = 0;
    for (int i = 1; i <= n; i++) {
      if (s[i] - '1') continue;
      bool tmp = 0;
      if (i > x && ans[i - x]) tmp = 1;
      if (i + x <= n && ans[i + x]) tmp = 1;
      if (!tmp) flg = 1;
    }
    if (flg)
      printf("-1\n");
    else {
      for (int i = 1; i <= n; i++) printf("%d", ans[i]);
      puts("");
    }
  }
  return 0;
}
