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
import java.util.Arrays;
import java.util.Comparator;
import java.util.Scanner;

import javax.lang.model.util.ElementScanner6;

public class Solution {
    static int[] col;
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int t = in.nextInt();
        in.nextLine();
        for (int z = 0; z < t; z++)
        {
            int n = in.nextInt(), m = in.nextInt(), a = in.nextInt(), b = in.nextInt();
            if (n * a != m * b)
            {
                System.out.println("NO");
            }
            else
            {
                System.out.println("YES");
            
            int[][] ans = new int[n][m];
            col = new int[m];
            int shift = 0;
            for (shift = 1; shift < m; shift++)
            {
                if (shift * n % m == 0)
                {
                    break;
                }
            }
            for (int i = 0, dx = 0; i < n; i++, dx += shift) {
                for (int j = 0; j < a; j++) {
                    ans[i][(j + dx) % m] = 1;
                }
            }
            String s = "";
            for (int i = 0; i < n; i++)
            {
                for (int j = 0; j < m; j++)
                {
                    s += (ans[i][j]);
                }
                s += "\n";
            }
            System.out.printf(s);
        }
        }

        in.close();
    }
}
    