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 double PI = 3.141592653589793238463;
const long long int MOD = 1000000007;
list<long long> ls;
vector<long long> v, v1, v2, vv;
set<long long> s;
map<long long, long long> mp;
stack<char> st;
stack<long long> sti;
inline long long fp(long long bs, long long pw) {
  long long res = 1;
  while (pw > 0) {
    if (pw % 2 == 1) {
      res = (res * bs) % MOD;
    }
    bs = (bs * bs) % MOD;
    pw = pw / 2;
  }
  return res;
}
long long pp(long long bs, unsigned long long pw) {
  long long ans = 1;
  while (pw > 0) {
    if (pw & 1) ans = ans * bs;
    pw = pw >> 1;
    bs = bs * bs;
  }
  return ans;
}
bool ff22(const pair<long long, long long> &a,
          const pair<long long, long long> &b) {
  return (a.second < b.second);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t, n, m, k, h, p, i, j, a, b, c, x, V, ind, sz,
      ans = 0, ans2, mx, sum = 0, ct = 0, ti, q, temp, t2, mn, ptr, ct1, ct2,
      cc, ln, diff;
  double kh = 4;
  bool fl;
  string s, sub, sorted, anss;
  cin >> t;
  while (t--) {
    ct = 0;
    cin >> s;
    n = s.length();
    sub = "";
    for (i = 0; i < n; i++) sub.push_back('1');
    cin >> x;
    for (i = 0; i < n; i++) {
      long long ms = i - x;
      long long pl = i + x;
      if (s[i] == '0') {
        if (ms >= 0) {
          sub[ms] = '0';
        }
        if (pl < n) {
          sub[pl] = '0';
        }
      }
    }
    for (i = 0; i < n; i++) {
      long long ms = i - x;
      long long pl = i + x;
      if (s[i] == '1') {
        if ((ms >= 0 && sub[ms] == '1') || (pl < n && sub[pl] == '1'))
          continue;
        else {
          ct++;
          break;
        }
      } else {
        if ((ms >= 0 && sub[ms] == '1') || (pl < n && sub[pl] == '1')) {
          ct++;
          break;
        }
      }
    }
    if (ct > 0)
      cout << -1 << endl;
    else
      cout << sub << endl;
  }
  return 0;
}
