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 inf = 0x3f3f3f3f;
void err() { cout << "\033[39;0m" << endl; }
template <class T, class... Ts>
void err(const T& arg, const Ts&... args) {
  cout << arg << " ";
  err(args...);
}
const int N = 50 + 5;
int T, n, m, a, b;
char s[N][N];
int col[N];
int main() {
  cin >> T;
  while (T--) {
    cin >> n >> m >> a >> b;
    if (n * a != m * b) {
      puts("NO");
      continue;
    }
    puts("YES");
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= m; j++) {
        s[i][j] = '0';
      }
    }
    memset(col, 0, sizeof col);
    for (int i = 1; i <= n; i++) {
      int cnt = a;
      while (cnt--) {
        int Min = n, pos = 0;
        for (int j = 1; j <= m; j++) {
          if (col[j] < Min) {
            Min = col[j];
            pos = j;
          }
        }
        col[pos]++;
        s[i][pos] = '1';
      }
    }
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= m; j++) {
        putchar(s[i][j]);
      }
      puts("");
    }
  }
  return 0;
}
