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 INF = 0x3f3f3f3f;
const int MAX_N = 5e5 + 20;
const int MAXM = 120010;
const long long MOD = 998244353;
const int MAXN = 2e5 + 20;
int n;
int x;
char s[MAXN];
char t[MAXN];
bool solve() {
  for (int i = 1; i <= n; i++) t[i] = '?';
  for (int i = 1; i <= n; i++) {
    if (s[i] == '0') {
      if (i - x >= 1 && t[i - x] == '1') return 0;
      if (i + x <= n) t[i + x] = '0';
      if (i - x >= 1 && t[i - x] == '?') t[i - x] = '0';
    } else {
      if (i - x >= 1 && t[i - x] == '1') {
        if (i + x <= n) t[i + x] = '?';
      } else {
        if (i - x >= 1 && t[i - x] == '?') {
          t[i - x] = '1';
          if (i + x <= n) t[i + x] = '?';
        } else if (i + x <= n)
          t[i + x] = '1';
        else
          return 0;
      }
    }
  }
  return 1;
}
int main() {
  int test;
  scanf("%d", &test);
  while (test--) {
    scanf("%s", s + 1);
    scanf("%d", &x);
    n = strlen(s + 1);
    if (solve()) {
      for (int i = 1; i <= n; i++) {
        if (t[i] != '?')
          printf("%c", t[i]);
        else
          printf("%c", '1');
      }
      printf("\n");
    } else {
      printf("-1\n");
    }
  }
}
