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 collections import Counter
from collections import defaultdict
from collections import deque
import math
import heapq
import sys
input = sys.stdin.readline
from  bisect import *
rs = lambda: input().strip()
ri = lambda: int(input())
rl = lambda: list(map(int, input().split()))
rls= lambda: list(map(str, input().split()))


def check(a):
    for i in range(0,r):
        if(a[i].count(1)!=n):
            return 0
    return 1

t=int(input())
for _ in range(0,t):
    r,c,n,m=rl()
    a=[]
    col=[0]*c
    for i in range(0,r):
        a.append([0]*c)
    x=0
    for i in range(0,r):
        for j in range(x,x+n):
            a[i][j%c]=1
            col[j%c]+=1
        #print(x,"ini")
        x=(x+n)%c
        #print(x)
    f=1
    d=col
    for i in range(0,c):
        if(d[i]>m):
            f=0
            break
        else:
            x=m-d[i]
        for j in range(0,r):
            if (x == 0):
                break
            if(a[j][i]==0):
                a[j][i]=1
                x=x-1
        if(x!=0):
            f=0
            break
    if(f and check(a)):
        print("YES")
        for i in range(0,r):
            for j in range(0,c):
                print(a[i][j],end="")
            print()
    else:
        print("NO")







