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 os
### START FAST IO ###
os_input = os.read(0, int(1e7)).split()
os_input_pos = -1
def cin():
    global os_input_pos
    os_input_pos += 1
    return os_input[os_input_pos]
# cout: os.write(1, "\n".encode().join(res))
#### END FAST IO ####

from math import gcd

T = int(cin())
res = []
while T:
    T -= 1
    n, m, a, b = [int(cin()) for i in range(4)]
    if n*a != m*b:
        res.append("NO")
        continue

    res.append("YES")
    g = [['0' for j in range(m)] for i in range(n)]
    ma = gcd(m, a)
    nb = gcd(n, b)
    a //= ma
    b //= nb
    m //= ma
    n //= nb
    for i in range(nb):
        for j in range(ma):
            for ii in range(n):
                for jj in range(a):
                    g[i*n + ii][j*m + (ii + jj) % m] = '1'
    res.append("\n".join([''.join(row) for row in g]))

os.write(1, "\n".join(res).encode())
