You are given four positive integers n, m, a, b (1 ≤ b ≤ n ≤ 50; 1 ≤ a ≤ m ≤ 50). Find any such rectangular matrix of size n × m that satisfies all of the following conditions:

  * each row of the matrix contains exactly a ones; 
  * each column of the matrix contains exactly b ones; 
  * all other elements are zeros. 



If the desired matrix does not exist, indicate this.

For example, for n=3, m=6, a=2, b=1, there exists a matrix satisfying the conditions above:

$$$ \begin{vmatrix} 0 & 1 & 0 & 0 & 0 & 1 \\\ 1 & 0 & 0 & 1 & 0 & 0 \\\ 0 & 0 & 1 & 0 & 1 & 0 \end{vmatrix} $$$

Input

The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.

Each test case is described by four positive integers n, m, a, b (1 ≤ b ≤ n ≤ 50; 1 ≤ a ≤ m ≤ 50), where n and m are the sizes of the matrix, and a and b are the number of ones for rows and columns, respectively.

Output

For each test case print:

  * "YES" (without quotes) and the required matrix (if there are several answers, print any) if it exists, or 
  * "NO" (without quotes) if it does not exist. 



To print the matrix n × m, print n rows, each of which consists of m numbers 0 or 1 describing a row of the matrix. Numbers must be printed without spaces.

Example

Input


5
3 6 2 1
2 2 2 1
2 2 2 2
4 4 2 2
2 1 1 2


Output


YES
010001
100100
001010
NO
YES
11
11
YES
1100
1100
0011
0011
YES
1
1
#include <bits/stdc++.h>
using namespace std;
class Compare {
 public:
  bool operator()(
      const pair<long long int, pair<long long int, long long int>> &p1,
      const pair<long long int, pair<long long int, long long int>> &p2) const {
    if (p1.first > p2.first)
      return true;
    else if (p1.first == p2.first) {
      if ((p1.second.first) < (p2.second.first)) return true;
    }
    return false;
  }
};
void solve() {
  long long int t;
  cin >> t;
  while (t--) {
    long long int n, m, a, b;
    cin >> n >> m >> a >> b;
    if (n * a == m * b) {
      cout << "YES\n";
      long long int d;
      for (long long int i = 1; i <= m; i++) {
        if ((i * n) % m == 0) {
          d = i;
          break;
        }
      }
      vector<long long int> pos;
      for (long long int i = 1; i <= a; i++) pos.push_back(i);
      vector<vector<bool>> v(n + 1, vector<bool>(m + 1, 0));
      for (long long int i = 1; i <= n; i++) {
        for (long long int j = 0; j < pos.size(); j++) {
          v[i][pos[j]] = 1;
          pos[j] = (pos[j] + d) % m;
          if (pos[j] == 0) pos[j] = m;
        }
        for (long long int j = 1; j <= m; j++) cout << v[i][j];
        cout << '\n';
      }
    } else
      cout << "NO\n";
  }
}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}
