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
.....
.###.
.....
.....
import sys

n, k = map(int, next(sys.stdin).rstrip().split())

max_hotels = (n - 2) * 2

if k > max_hotels:
    print("NO")
    sys.exit()

def form_line(k):
    s = list(('#' * k).center(n, '.'))
    s[n // 2] = '.'
    return ''.join(s)

print("YES")

if k % 2 == 1:
    print('.' * n)
    if k > n - 2:
        print('.%s.' % ('#' * (n - 2)))
        print(form_line(k - (n - 2) + 1))
    else:
        print(('#' * k).center(n, '.'))
        print('.' * n)
    print('.' * n)
else:
    xs = []
    xs.append('....')
    for _ in range(n - 2):
        if k == 0:
            xs.append('....')
        else:
            xs.append('.##.')
            k -= 2
    xs.append('....')

    for p in range(4):
        print(''.join(x[p] for x in xs))
