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 n, m, a, b;
int g[55][55];
signed main() {
  int t;
  cin >> t;
  while (t--) {
    cin >> n >> m >> a >> b;
    memset(g, 0, sizeof g);
    for (int j = 1; j <= m; ++j) {
      for (int i = 1; i <= b; ++i) g[i][j] = 1;
    }
    for (int i = 1; i <= n; ++i) {
      int cnt = 0;
      for (int j = 1; j <= m; ++j) cnt += g[i][j];
      if (cnt == a)
        continue;
      else if (cnt < a) {
        puts("NO");
        goto loop;
      } else {
        for (int j = 1; j <= m && cnt > a; ++j) {
          if (g[i][j] == 0) continue;
          for (int k = i + 1; k <= n; ++k) {
            if (g[k][j] == 0) {
              cnt--;
              g[k][j] = 1;
              g[i][j] = 0;
              break;
            }
          }
        }
        if (cnt > a) {
          puts("NO");
          goto loop;
        }
      }
    }
    puts("YES");
    for (int i = 1; i <= n; ++i) {
      for (int j = 1; j <= m; ++j) cout << g[i][j];
      cout << '\n';
    }
  loop:
    continue;
  }
  return 0;
}
