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 main() {
  long long int test_case;
  cin >> test_case;
  while (test_case--) {
    long long int n, m, a, b;
    cin >> n >> m >> a >> b;
    if (n * a != m * b)
      cout << "NO" << endl;
    else {
      cout << "YES" << endl;
      long long int i, j, t;
      long long int temp1;
      for (i = 1; i < 100; i++) {
        if ((i * n) % m == 0) {
          temp1 = i;
          break;
        }
      }
      vector<vector<int> > vec1;
      int prev_mark = 0;
      for (i = 0; i < n; i++) {
        vector<int> vec2(m, 0);
        if (i == 0) {
          for (j = prev_mark; j < prev_mark + a; j++) {
            vec2[j] = 1;
          }
          if (prev_mark + temp1 >= m) {
            prev_mark = m - prev_mark - temp1;
          } else {
            prev_mark += temp1;
          }
        } else {
          for (j = prev_mark; j < min(m, prev_mark + a); j++) {
            vec2[j] = 1;
          }
          if (min(m, prev_mark + a) == m) {
            for (j = 0; j < prev_mark + a - m; j++) {
              vec2[j] = 1;
            }
          }
          if (prev_mark + temp1 >= m) {
            prev_mark = m - prev_mark - temp1;
          } else {
            prev_mark += temp1;
          }
        }
        vec1.push_back(vec2);
      }
      for (i = 0; i < n; i++) {
        for (j = 0; j < m; j++) {
          cout << vec1[i][j];
        }
        cout << endl;
      }
    }
  }
}
