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 numpy as np
def find_matrix():
    n,m,a,b = list(map(int,input().split()))
 
    if n*a != m*b:
        return 'NO'

    #check square matrix
    # if n == m:
    #     if a == b:
    #         # print('YES')
    #         matrix = []
    #         matrix_in = []
    #         for i in range(m):
    #             matrix_in.append(1)
    #         for i in range(n):
    #             matris = matrix_in.copy()
    #             matrix.append(matris)
    #         # matrix = np.ones((n,m),dtype=np.int8)
    #         zero = m-a #1
    #         active_column = []
    #         for i in range(1,zero+1):
    #             active_column.append(n-i)
    #         # print(active_column)
    #         for i in range(n):
    #             for k in active_column:
    #                 matrix[i][k] = 0
    #             for j in range(len(active_column)):
    #                 active_column[j] -= 1
    #                 if active_column[j] == -1:
    #                     active_column[j] = n-1
    #         out_return = 'YES'
    #         for i in matrix:
    #             out = ''
    #             for j in i:
    #                 out += str(j) 
    #             out_return += '\n'+out
    #         return out_return
    
    # if (n%m == 0 or m%n == 0) and (a%b == 0 or b%a == 0) and (a!=b):
        # if (n>m and b>a) or (m>n and a>b): 
    matrix = []
    matrix_in = []
    for i in range(m):
        matrix_in.append(1)
    for i in range(n):
        matris = matrix_in.copy()
        matrix.append(matris)
    zero = m-a #1
    # active_range = []
    start = 0
    end = zero

    # print(start,end)
    # print(active_range)
    for i in range(n):
        for j in range(start,end):
            j %= m
            matrix[i][j] = 0
        start += zero
        end += zero
    out_return = 'YES'
    for i in matrix:
        out = ''
        for j in i:
            out += str(j) 
        out_return += '\n'+out
    return out_return
 
    # elif n%m == 0 or m%n == 0:
    #     if a%b == 0 or b%a == 0:
 
    #         # do something
    #     else:
    #         print('NO')
    # else:
    #     print('NO')
t = int(input())
output = []
for i in range(t):
    output.append(find_matrix())
for i in output:
    print(i)