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
.....
.###.
.....
.....
n,k=list(map(int, input().split(' ')))
def print_res(lst):
    for i in lst:
        print(''.join(i))
res=[['.']*n] + [['.']*n] + [['.']*n] + [['.']*n]


if k==0:
    print('YES')
    print_res(res)
elif k>(n-2)*2:
    print('NO')
elif k==n-2:
    print('YES')
    res[1]=['.']+['#'*k]+['.']
    # res[1][0]='.'
    # res[1][-1]='.'
    print_res(res)
elif k%2==0:
    print("YES")
    for i in range(k//2):
        res[1][i+1]='#'
        res[2][i+1]='#'
    print_res(res)
elif k<(n-2):
    print("YES")
    res[1]=['.'*((n-k)//2)]+['#'*(k)]+['.'*((n-k)//2)]
    print_res(res)
elif k>(n-2):
     print("YES")
     res[1]=['.']+['#'*(n-2)]+['.']
     x=(k-n+2)//2
     for i in range(x):
         res[2][i+1]='#'
         res[2][-i-2]='#'
     # res[2]=['.']+['#'*()]+['.']+['#'*((k-n+2)//2)]+['.']
     print_res(res)
else:
    print('NO')
