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
t=int(input())
for i in range(t):
    n,m,a,b=map(int,input().split())
    lst=[]
    for i in range(n):
        arr=[]
        for j in range(m):
            arr+=[0]
        lst+=[arr]
    #lst[1][2]=1
    #print(lst)
    j,k=0,0
    x=0
    while(j<n):
        while(k<m):
            if(x==a):
                x=0
            while(x<a):
                if(k<m):
                    lst[j][k]=1
                    k+=1
                else:
                    break
                x+=1
            if(x==a):
                j+=1
            if(j>=n):
                break
        k=0
    flag=0
    for c in range(m):
        count=0
        for d in range(n):
            if(lst[d][c]==1):
                count+=1
        if(count!=b):
            flag=1
            print("NO")
            break
    if(flag==0):
        print("YES")
        for c in range(n):
            for d in range(m):
                print(lst[c][d], end="")
            print()
