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 sys
import math
# sys.setrecursionlimit(10**6)

def get_ints(): return map(int, sys.stdin.readline().strip().split())
def get_array(): return list(map(int, sys.stdin.readline().strip().split()))

def printspx(*args): return print(*args, end="")
def printsp(*args): return print(*args, end=" ")
def printchk(*args): return print(*args, end=" \ ")

MODPRIME = int(1e9+7); BABYMOD = 998244353;

if __name__ == "__main__":
    # sys.stdin = open("input.txt","r")  # <<<  ^o^  Comment this line  ^_^  >>>
    for _testcases_ in range(int(input())):
        n, m, a, b = get_ints()
        if m/a != n/b :
            print("NO")
        else:
            print("YES")
            matr = [[0]*m for x in range(n)]
            fac = math.ceil(m/a)
            for i in range(n):
                for j in range(a):
                    matr[i][(j+(i*a))%m] = 1
            for i in range(n):
                for j in range(m):
                    printspx(matr[i][j])
                print()
                    
            


# #############################################################################
'''
THE LOGIC AND APPROACH IS BY ME @luctivud ( UDIT GUPTA )
SOME PARTS OF THE CODE HAS BEEN TAKEN FROM WEBSITES LIKE::
(I Own the code if no link is provided here or I may have missed mentioning it)
>>> DO NOT PLAGIARISE.
TESTCASES:
'''