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
.....
.###.
.....
.....
from bisect import bisect_left, bisect_right

def tuple_input(type):
    return map(type, input().strip().split())

def print_list(s):
    print(''.join(map(str, s)))


'''
'''
n, k = tuple_input(int)
res = list()
for i in range(4):
    res.append(['.'] * n)
if k == 1:
    res[1][n // 2] = '#'
elif k == 3:
    res[1][n // 2 - 1] = '#'
    res[1][n // 2    ] = '#'
    res[1][n // 2 + 1] = '#'
else:
    for i in range(1, 3):
        for j in range(1, ((k + 1) // 2) + 1):
            res[i][j] = '#'
    if k % 2 != 0:
        res[1][2] = '.'
print('YES')
for e in res:
    print(''.join(e))