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;
int dx[8] = {0, 0, 1, -1, 1, -1, 1, -1};
int dy[8] = {1, -1, 0, 0, 1, -1, -1, 1};
int arr[55][55];
int main() {
  ios_base::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    int n, m, a, b;
    cin >> n >> m >> a >> b;
    if (n * a != m * b) {
      cout << "NO" << endl;
      continue;
    }
    cout << "YES\n";
    int c = 0, x = 0;
    for (int i = 0; i < n && c < n * a; ++i) {
      for (int j = 0; j < a && c < n * a; c++, ++j) {
        arr[i][x] = 1;
        x = (x + 1) % m;
      }
    }
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < m; ++j) {
        cout << arr[i][j];
      }
      cout << endl;
    }
    for (int i = 0; i < n; ++i)
      for (int j = 0; j < m; ++j) arr[i][j] = 0;
  }
  return 0;
}
