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, math
from sys import stdin, stdout
rem = 10 ** 9 + 7
inf=10**9
sys.setrecursionlimit(10**5)
from heapq import heappush,heappop,heapify
from collections import deque
take = lambda: map(int, stdin.readline().split())


def check(new,arr,m):
    done=True
    #print new,arr
    for p in arr:
        count=0
        for i in range(m):
            if new[i]!=p[i]:
                count+=1
        if count>1:
            done=False
            break

    return done


t= input()
for q in range(t):
    n,m,a,b=take()
    arr=[['0' for i in range(m)] for j in range(n)]
    if a*n == m*b and n>=b and m>=a:
        print 'YES'
    else:
        print 'NO'
        continue

    prevcol=0
    fin=[]
    for i in range(n):
        for k in range(a):
            if prevcol==m:
                prevcol=0
            arr[i][prevcol]='1'
            prevcol+=1
        fin.append(''.join(arr[i]))
    #print fin
    print '\n'.join(fin)




