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
def main():
    
    t=int(input())
    for _a in range(t):
        n,m,a,b=readIntArr()
        
#        Number of 1s must be equal when calculated by row and column.
#        So n*a==m*b.
#        If they are equal, then as long as I put a 1s per row and 
#        evenly distribute them by column, the solution will work. To
#        evenly distribute them by column, offset the previous row when
#        constructing the current row by a common difference d=a, since each
#        row will cause an inbalance in the [offset, offset+a) column-group. The
#        final offset will be n*a which is a multiple of m, which guarantees that it ends
#        at the completion of a cycle of column-groups.
        
        if n*a==m*b:
            print('YES')
            matrix=[[0 for _ in range(m)] for __ in range(n)]
            d=a
            start=0
            for row in range(n):
                i=start
                for __ in range(a):
                    matrix[row][i%m]=1
                    i+=1
                start+=d
            multiLineArrayOfArraysPrint(matrix)
        else:
            print('NO')
            
    
    return
 
import sys
input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)
#import sys
#input=lambda: sys.stdin.readline().rstrip("\r\n") #FOR READING STRING/TEXT INPUTS.
 
def oneLineArrayPrint(arr):
    print(' '.join([str(x) for x in arr]))
def multiLineArrayPrint(arr):
    print('\n'.join([str(x) for x in arr]))
def multiLineArrayOfArraysPrint(arr):
    print('\n'.join([''.join([str(x) for x in y]) for y in arr]))
 
def readIntArr():
    return [int(x) for x in input().split()]
 
inf=float('inf')
MOD=10**9+7
 
main()