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
.....
.###.
.....
.....
"""Codeforces P980B. Marlin
    (http://codeforces.com/problemset/problem/980/B)

Problem tags: constructive algorithms

Hint: A symmetric grid in either horizontal or vertical direction can be an 
    answer.
Note: Be cautious that "hotels should not be placed on the border cells".
Time Complexity: O(n).  (n <= 99)
"""

def main():
    n, k = [int(x) for x in input().split()]

    l1 = ['.'] * n
    l2 = ['.'] * n
    if k % 2 == 1:
        h1 = min(k, n - 2)
        h2 = (k - h1) // 2
        l1[(n - h1) // 2: (n - h1) // 2 + h1] = ['#'] * h1
        l2[1 : h2 + 1] = ['#'] * h2
        l2[-2 : -h2 - 2: -1] = ['#'] * h2
    else:
        l1[1 : k // 2 + 1] = ['#'] * (k // 2)
        l2[1 : k // 2 + 1] = ['#'] * (k // 2)
        
    print('YES')            
    print('.' * n)
    print(''.join(l1))
    print(''.join(l2))
    print('.' * n)
    
    
if __name__ == '__main__':
    main()    
