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
.....
.###.
.....
.....
a = input()

n, k = a.split(' ')
n = int(n)
k = int(k)

if (2*(n-2) < k):
    print('NO')

else:
    print('YES')
    kiekeil = k // (n - 2)
    kiekpask = k % (n - 2)
    for i in range(0, n):
        print('.', end='')
    print()
    for i in range(0, kiekeil):
        print('.', end='')
        for i in range(1, n-1):
            print('#', end='')
        print('.', end='')    
        print()
    if(kiekpask % 2 == 0):
       print('.', end='')
       for i in range(0, kiekpask//2):
           print('#', end='')
       for i in range(0, n-2-kiekpask):
           print('.', end='')
       for i in range(0, kiekpask//2):
           print('#', end='')
       print('.', end='')
       print()
       if (kiekeil == 0):
            for i in range(0, n):
               print('.', end='') 
            print()
       if(kiekpask != 0 or kiekeil != 2):
           for i in range(0, n):
               print('.', end='')            
    else:
       print('.', end='')
       for i in range(1, (n-2-kiekpask)//2+1):
           print('.', end='')
       for i in range(0, kiekpask):
           print('#', end='')
       for i in range(1, (n-2-kiekpask)//2+1):
           print('.', end='')
       print('.', end='')
       print()
       if (kiekeil == 0):
            for i in range(0, n):
               print('.', end='') 
            print()
       for i in range(0, n):
           print('.', end='')
