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
n = int(input())


def r_s(x, j, n):
    sum = 0
    for i in range(n):
        sum += x[i][j]
    return sum


def o_z(x, j1, j2, n):
    for i in range(n):
        if x[i][j1] == 1 and x[i][j2] == 0:
            return i


for _ in range(n):
    n, m, a, b = [int(s) for s in input().split()]
    if n * a != m * b:
        print('NO')
    else:
        print('YES')
        x = [[1 if i < a else 0 for i in range(m)] for j in range(n)]
        sums = [n] * a + [0] * (m - a)
        for j1 in range(m):
            sum1 = sums[j1]
            if sum1 > b:
                for t in range(sum1 - b):
                    for j2 in range(1, m):
                        sum2 = sums[j1 - j2]
                        if sum2 < b:
                            i1 = o_z(x, j1, j1 - j2, n)
                            x[i1][j1], x[i1][j1 - j2] = x[i1][j1 - j2], x[i1][j1]
                            sums[j1], sums[j1 - j2] = sums[j1] - 1, sums[j1 - j2] + 1
                            break
        for i in range(n):
            print(*x[i], sep='')