The city of Fishtopia can be imagined as a grid of 4 rows and an odd number of columns. It has two main villages; the first is located at the top-left cell (1,1), people who stay there love fishing at the Tuna pond at the bottom-right cell (4, n). The second village is located at (4, 1) and its people love the Salmon pond at (1, n).

The mayor of Fishtopia wants to place k hotels in the city, each one occupying one cell. To allow people to enter the city from anywhere, hotels should not be placed on the border cells.

A person can move from one cell to another if those cells are not occupied by hotels and share a side.

Can you help the mayor place the hotels in a way such that there are equal number of shortest paths from each village to its preferred pond?

Input

The first line of input contain two integers, n and k (3 ≤ n ≤ 99, 0 ≤ k ≤ 2×(n-2)), n is odd, the width of the city, and the number of hotels to be placed, respectively.

Output

Print "YES", if it is possible to place all the hotels in a way that satisfies the problem statement, otherwise print "NO".

If it is possible, print an extra 4 lines that describe the city, each line should have n characters, each of which is "#" if that cell has a hotel on it, or "." if not.

Examples

Input

7 2


Output

YES
.......
.#.....
.#.....
.......


Input

5 3


Output

YES
.....
.###.
.....
.....
def main():
	n, k = [int(x) for x in input().split()]
	if 2*(n-2) < k:
		print('NO')
	else:
		print('YES')
		mat = [['.']*n for x in range(4)]
		if k % 2 == 0:
			for i in range(1,k//2+1):
				mat[1][i] = '#'
				mat[2][i] = '#'
			for i in range(4):
				print(''.join(mat[i]))
		else:
			c = k
			print('.'*n)
			for i in range(1,3):
				if c >= n-2:
					print(('#' * (n-2)).center(n,'.'))
					c -= n-2
				elif c>0:
					if c % 2 == 1:
						print(('#' * c).center(n, '.'))
					else:
						print('.'+('.' * (n-c-2)).center(n-2, '#')+'.') 
					c -= n-2
				else:print('.'*n)
			print('.'*n)
if __name__ == '__main__':
	main()