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
# -*- coding: utf-8 -*-
"""
Created on Sat Jun 27 04:15:22 2020

@author: Dark Soul
"""

n=input('')
n=int(n)
str=[]

for i in range(n):
    entries = list(map(int, input().split()))
    str.append(entries)
for i in str:
    row=i[0]
    column=i[1]
    a=i[2]
    b=i[3]
    if row*a!=column*b:
        print('NO')
    else:
        print('YES')
        dis=0
        col_ind=0
        row_ind=0
        final_matrix=[]
        
        for k in range(1,column+1,1):
            if((k*row)%column==0):
                dis=k;
                break
            
        for l in range(row):
            
            mat=['0']*column
            
            for j in range(a):
                
                col_ind=(row_ind+j)%column
                mat[col_ind]='1'
                
            row_ind=row_ind+dis
            final_matrix.append(mat)
        for i in range(row):
            print(''.join(final_matrix[i]))
        
        
        
            
        
    