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
from sys import stdin, stdout
import heapq
import cProfile
from collections import Counter, defaultdict, deque
from functools import reduce
import math


def get_int():
    return int(stdin.readline().strip())


def get_tuple():
    return map(int, stdin.readline().split())


def get_list():
    return list(map(int, stdin.readline().split()))


def solve():
    n,m,a,b = map(int, input().strip().split())
    matrix = [[0 for _ in range(m)] for _ in range(n)]
    if n*a!=m*b: print("NO")
    else:
        ans = "YES"
        j = 0
        for i in range(n):
            tempb,tempa = b,a
            f = 0
            while tempa>0 and f!=2:
                    matrix[i][j] = 1
                    tempa -= 1
                    j = (j+1)%m
                    if j==i%m:
                        f += 1
        print(ans)
        if ans=="YES":
            for i in range(n):
                print("".join(str(e) for e in matrix[i]))






def main():
    solve()


TestCases = True

if TestCases:
    for i in range(get_int()):
        main()
else:
    main()
