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.io.*;

public class G
{
    public static void main(String args[])throws Exception
    {
        BufferedReader bu=new BufferedReader(new InputStreamReader(System.in));
        StringBuilder sb=new StringBuilder();
        int t=Integer.parseInt(bu.readLine());
        while(t-->0)
        {
            String s[]=bu.readLine().split(" ");
            int n=Integer.parseInt(s[0]),m=Integer.parseInt(s[1]),a=Integer.parseInt(s[2]),b=Integer.parseInt(s[3]);
            int ans[][]=new int[n][m],i,j,d,sh=0;
            if(n*a!=m*b) {sb.append("NO\n"); continue;}
            sb.append("YES\n");
            for(i=1;i<m;i++)
            if(n*i%m==0) {sh=i; break;}
            for(i=0,d=0;i<n;i++,d+=sh)
            for(j=0;j<a;j++)
                ans[i][(j+d)%m]=1;
            for(i=0;i<n;i++)
            {for(j=0;j<m;j++)
            sb.append(ans[i][j]);
            sb.append("\n");}
        }
        System.out.print(sb);
    }
}