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;
const int maxm = 2e3 + 23;
const int maxn = 2e5 + 35;
int aa[100][100];
int main() {
  int T, a, b, n, m;
  cin >> T;
  while (T--) {
    cin >> n >> m >> a >> b;
    memset(aa, 0, sizeof(aa));
    if (a * n != b * m) {
      puts("NO");
      continue;
    }
    int s = 1;
    for (int i = 1; i <= n; ++i) {
      for (int j = 1; j <= a; ++j) {
        aa[i][s] = 1;
        s += 1;
        if (s > m) s = 1;
      }
    }
    puts("YES");
    for (int i = 1; i <= n; ++i) {
      for (int j = 1; j <= m; ++j) cout << aa[i][j];
      puts("");
    }
  }
  return 0;
}
