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.*;
import java.io.*;
public class Solution implements Runnable{
    FastScanner sc;
    PrintWriter pw;
    final class FastScanner {
        BufferedReader br;
        StringTokenizer st;
 
        public FastScanner() {
            try {
                br = new BufferedReader(new InputStreamReader(System.in));
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                e.printStackTrace();
            }
        }
 
        public long nlo() {
            return Long.parseLong(next());
        }
 
        public String next() {
            if (st.hasMoreTokens()) return st.nextToken();
            try {
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                e.printStackTrace();
            }
            return st.nextToken();
        }
 
        public int ni() {
            return Integer.parseInt(next());
        }
 
        public String nli() {
            String line = "";
            if (st.hasMoreTokens()) line = st.nextToken();
            else try {
                return br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            while (st.hasMoreTokens()) line += " " + st.nextToken();
            return line;
        }
 
        public double nd() {
            return Double.parseDouble(next());
        }
    }
    public static void main(String[] args) throws Exception
    {
        new Thread(null,new Solution(),"codeforces",1<<28).start();
    }
    public void run()
    {
        sc=new FastScanner();
        pw=new PrintWriter(System.out);
        solve();
        pw.flush();
        pw.close();
    }
    public long gcd(long a,long b)
    {
        return b==0L?a:gcd(b,a%b);
    }
    public long ppow(long a,long b,long mod)
    {
        if(b==0L)
        return 1L;
        long tmp=1;
        while(b>1L)
        {
            if((b&1L)==1)
            tmp*=a;
            a*=a;
            a%=mod;
            tmp%=mod;
            b>>=1;
        }
        return (tmp*a)%mod;
    }
    public  int gcd(int x,int y)
    {
        return y==0?x:gcd(y,x%y);
    }
    //////////////////////////////////
    /////////////  LOGIC  ///////////
    ////////////////////////////////
    public void solve()
    {
        int t=sc.ni();
        while(t-->0)
        {
            int n=sc.ni();
            int m=sc.ni();
            int[][] arr=new int[n][m];
            int a=sc.ni();
            int b=sc.ni();
            int res=0;
            if(a*n!=b*m)
            res=-1;
            else
            {
                if(m>=n)
                {int r=0;
                    for(int j=0;j<m;j++)
                    {
                        for(int p=0;p<b;p++)
                            arr[(r+p)%n][j]=1;
                        r=(r+b)%n;
                    }
                }
                else
                {
                    int c=0;
                    for(int r=0;r<n;r++)
                    {
                        for(int p=0;p<a;p++)
                        arr[r][(c+p)%m]=1;
                        c=(c+a)%m;
                    }
                }
            }
            if(res==0)
            {
                pw.println("YES");
                for(int i=0;i<n;i++)
                {
                    for(int j=0;j<m;j++)
                    pw.print(arr[i][j]);
                    pw.println();
                }
            }
            else
            pw.println("NO");
        }
    }
}