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
T = int(input())
for t in range(T):
    n, m, a, b = list(map(int, input().split()))
    if n * a != m * b:
        res = "NO"
    else:
        # print("n, m, a, b", n, m, a, b)
        matrix = [["0" for j in range(m)] for i in range(n)]
        res = "YES"
        steps = 0
        max_steps = n * a
        cur_col = 0
        cur_row = 0
        # print(max_steps)
        while steps < max_steps:
            # print(cur_row, cur_col, steps)
            matrix[cur_row][cur_col] = "1"
            cur_col += 1
            if cur_col == m:
                cur_col = 0
            steps += 1
            if (steps % a) == 0:
                cur_row += 1
            # res = "YES"
            # col_index = 0
            # cur_col = 0
            # total_col = 0
            # for i in range(n):
            #     if cur_col == b:
            #         col_index += a
            #         cur_col = 0
            #     for j in range(col_index, col_index + a):
            #         print("i, j", i, j)
            #         matrix[i][j] = "1"
            #     cur_col += 1
            # total_col += 1
    print(res)
    if res == "YES":
        for i in range(n):
            print("".join(matrix[i]))
