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[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
int dx8[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dy8[] = {1, -1, -1, 0, 1, -1, 0, 1};
int kx8[] = {1, 1, 2, 2, -1, -1, -2, -2};
int ky8[] = {2, -2, 1, -1, 2, -2, 1, -1};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long t, tt = 1;
  cin >> t;
  while (t--) {
    long long a, b, n, m;
    string s[55];
    cin >> n >> m >> a >> b;
    int cl[55] = {0};
    int rw[55] = {0};
    for (int i = 0; i < (n); i++)
      for (int j = 0; j < (m); j++) s[i].push_back('0');
    for (int i = 0; i < (n); i++) {
      for (int j = 0; j < (a); j++) {
        s[i][j] = '1';
        rw[i]++;
        cl[j]++;
      }
    }
    for (int i = 0; i < (n); i++)
      for (int j = 0; j < (m); j++) {
        if (cl[j] > b && s[i][j] == '1') {
          for (int k = 0; k < m; k++) {
            if (s[i][k] == '0' && cl[k] < b) {
              s[i][j] = '0';
              s[i][k] = '1';
              cl[j]--;
              cl[k]++;
              break;
            }
          }
        }
      }
    bool flg = 1;
    for (int i = 0; i < (n); i++)
      if (rw[i] != a) flg = 0;
    for (int i = 0; i < (m); i++)
      if (cl[i] != b) flg = 0;
    if (flg) {
      cout << "YES" << endl;
      for (int i = 0; i < (n); i++) cout << s[i] << endl;
    } else
      cout << "NO" << endl;
  }
  return 0;
}
