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 = 1e9 + 1000;
const long long INF64 = 1e18;
const int N = 1001000;
const int MOD = 1e9 + 7;
const double EPS = 1E-9;
long long gcd(long long a, long long b) { return a == 0 ? b : gcd(b % a, a); }
int add(int first, int second, int m) {
  first += second;
  while (first >= m) first -= m;
  while (first < 0) first += m;
  return first;
}
int sub(int first, int second, int m) { return add(first, -second, m); }
int mul(long long first, long long second, int m) {
  first %= m;
  second %= m;
  return (first * 1ll * second) % m;
}
int binpow(int first, int second, int m) {
  int z = 1;
  while (second) {
    if (second & 1) z = mul(z, first, m);
    first = mul(first, first, m);
    second >>= 1;
  }
  return z;
}
int divide(int first, int second, int m) {
  return mul(first, binpow(second, m - 2, m), m);
}
long long binpow(long long first, long long second) {
  long long z = 1;
  while (second) {
    if (second & 1) z *= first;
    first = first * first;
    second >>= 1;
  }
  return z;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    string s;
    int first;
    cin >> s >> first;
    string res(s.size(), ' ');
    int done = 0;
    int cur_idx = 0;
    int n = s.size();
    for (int i = 0; i < n; ++i) {
      if (s[i] == '0') {
        if (i - first >= 0 && i + first < n) {
          res[i - first] = res[i + first] = '0';
        } else if (i - first < 0 && i + first < n) {
          res[i + first] = '0';
        } else if (i - first >= 0 && i + first >= n) {
          res[i - first] = '0';
        }
      }
    }
    for (int i = 0; i < n; ++i) {
      if (res[i] == ' ') {
        res[i] = '1';
      }
    }
    bool ok = true;
    for (int i = 0; i < n; ++i) {
      if (i - first >= 0 && res[i - first] == '1') {
        ok &= (s[i] == '1');
      } else if (i + first < n && res[i + first] == '1') {
        ok &= (s[i] == '1');
      } else {
        ok &= (s[i] == '0');
      }
    }
    cout << (ok ? res : "-1") << endl;
  }
}
