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;
using namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
long long int gcd(long long int a, long long int b) {
  if (a < b)
    return gcd(b, a);
  else if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long int isPrime(long long int n) {
  long long int p = (long long int)sqrt(n);
  for (int i = (2); i <= (p); i += (1))
    if (n % i == 0) return 0;
  return 1;
}
long long int pow(long long int b, long long int e) {
  if (e == 0)
    return 1;
  else if (e % 2 == 0) {
    long long int a = pow(b, e / 2);
    return a * a;
  } else {
    long long int a = pow(b, e / 2);
    return b * a * a;
  }
}
long long int powm(long long int x, long long int y,
                   long long int m = 1000000007) {
  x = x % m;
  long long int res = 1;
  while (y) {
    if (y & 1) res = res * x;
    res %= m;
    y = y >> 1;
    x = x * x;
    x %= m;
  }
  return res;
}
long long int modInverse(long long int a, long long int m) {
  return powm(a, m - 2, m);
}
void solve() {
  string s;
  cin >> s;
  long long int x, n = s.length();
  cin >> x;
  vector<char> w(n, '1');
  for (int i = (0); i <= (n - 1); i += (1)) {
    if (i >= x and s[i] == '0') w[i - x] = '0';
    if (i + x <= n - 1 and s[i] == '0') w[i + x] = '0';
  }
  string c;
  for (int i = (0); i <= (n - 1); i += (1)) {
    if ((i < x or (w[i - x] == '0')) and (i + x >= n or (w[i + x] == '0')))
      c += '0';
    else
      c += '1';
  }
  if (c == s) {
    for (int i = (0); i <= (n - 1); i += (1)) cout << w[i];
    cout << '\n';
  } else
    cout << -1 << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int T;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}
