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;
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << ' ' << *it << " = " << a;
  err(++it, args...);
}
template <typename T, typename U>
inline void min_self(T& x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
inline void max_self(T& x, U y) {
  if (x < y) x = y;
}
template <class T1, class T2>
ostream& operator<<(ostream& out, pair<T1, T2> pair) {
  return out << "(" << pair.first << ", " << pair.second << ")";
}
template <class T>
ostream& operator<<(ostream& out, vector<T> vec) {
  out << "(";
  for (auto& v : vec) out << v << ", ";
  return out << ")";
}
template <class T>
ostream& operator<<(ostream& out, set<T> vec) {
  out << "(";
  for (auto& v : vec) out << v << ", ";
  return out << ")";
}
template <class L, class R>
ostream& operator<<(ostream& out, map<L, R> vec) {
  out << "(";
  for (auto& v : vec) out << "[" << v.first << ", " << v.second << "]";
  return out << ")";
}
template <class A, class B>
istream& operator>>(istream& in, pair<A, B>& a) {
  return in >> a.first >> a.second;
}
template <class A>
istream& operator>>(istream& in, vector<A>& a) {
  for (A& i : a) in >> i;
  return in;
}
long long XX[4] = {-1, 0, 1, 0};
long long YY[4] = {0, -1, 0, 1};
void solve() {
  string s;
  cin >> s;
  int n, x;
  cin >> x;
  n = s.length();
  vector<int> a(n, -1);
  for (int i = 0; i < n; i++) {
    if (s[i] == '0') {
      if (i >= x) a[i - x] = 0;
      if (i + x < n) a[i + x] = 0;
    }
  }
  for (int i = 0; i < n; i++) {
    if (s[i] == '1') {
      int cnt = 0;
      if (i >= x && a[i - x] == -1) {
        cnt++;
      }
      if (i + x < n && a[i + x] == -1) {
        cnt++;
      }
      if (cnt == 0) {
        cout << -1 << '\n';
        return;
      }
    }
  }
  for (int i : a) {
    if (i == -1)
      cout << '1';
    else
      cout << '0';
  }
  cout << '\n';
  return;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
