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
FAST_IO = 1
if FAST_IO:
    import io, sys, atexit
    rr = iter(sys.stdin.read().splitlines()).next
    sys.stdout = _OUTPUT_BUFFER = io.BytesIO()
    @atexit.register
    def write():
        sys.__stdout__.write(_OUTPUT_BUFFER.getvalue())
else:
    rr = raw_input
rri = lambda: int(rr())
rrm = lambda: map(int, rr().split())

####

from collections import deque, defaultdict as ddic, Counter
import itertools, heapq
MOD = 10 ** 9 + 7
YES, NO = "YES", "NO"

def solve(R, C, horiz_ones, vert_ones):
    A = [[0] * C for _ in xrange(R)]
    if R * horiz_ones != C * vert_ones:
        return
    for shift in range(C):
        count = [0] * C
        x = 0
        for r in xrange(R):
            for j in range(horiz_ones):
                count[(x + j) % C] += 1
            x += shift
            x %= C
        if all(v==vert_ones for v in count):
            # shift is good
            x = 0
            for r in xrange(R):
                for j in range(horiz_ones):
                    A[r][(x + j) % C] += 1
                x += shift
                x %= C
            return A
    return None
    
for _ttt in xrange(rri()):
    ans = solve(*rrm())
    if ans is None:
        print NO
    else:
        print YES
        for row in ans:
            print "".join(map(str, row))
