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=[int(x) for x in input().split()[:2]]
def pri(a,b):
    for i in range(b):
        print(a,end='')

def abd():
    if k%2==0:
        print('YES')
        for i in range(n):
            print('.',end='')
        print()
        for j in range(2):
            print('.', end='')
            for i in range(k//2):
                print('#', end='')
            for i in range(k//2,n-1):
                print('.', end='')
            print()
        for i in range(n):
            print('.',end='')
        print()
    else: # k is odd
        if k==1:
            print('YES')
            pri('.',n)
            print()
            pri('.',n//2)
            print('#',end='')
            pri('.',n//2)
            print()
            pri('.',n)
            print()
            pri('.',n)
            print()
            return
        if k==3:
            print('YES')
            pri('.',n)
            print()
            pri('.',n//2-1)
            print('###',end='')
            pri('.',n//2-1)
            print()
            pri('.',n)
            print()
            pri('.',n)
            print()
            return
        b=k//2
        a=b+1
        print('YES')
        pri('.',n)
        print()
        print('.',end='')
        pri('#',a)
        pri('.',n-a-1)
        print()
        print('.',end='')
        pri('#',b-1)
        print('.',end='')
        print('#',end='')
        pri('.',n-a-1)
        print()
        pri('.',n)
        print()
abd()