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=map(int,input().split())
ch1="."*n
"""LOOOOOOOOOOOOOOOOOOOOOOOOOL this is so easy man i want some juicy ratinng"""
if(k%2==0):
      print("YES")
      ch2="."+"#"*(k//2)+"."*(n-(k//2)-1)
      ch3="."+"#"*(k//2)+"."*(n-(k//2)-1)
      print(ch1)
      print(ch2)
      print(ch3)
      print(ch1)
else:
      if(k<=n-2):
            print("YES")
            ch2="."*((n-k)//2)+"#"*k+"."*((n-k)//2)
            ch3=ch1
            print(ch1)
            print(ch2)
            print(ch3)
            print(ch1)
      else:
            print("YES")
            p=k-n+2
            ch2="."+"#"*(n-2)+"."
            ch3="."+"#"*(p-1)+"."*(n-p-2)+"#"+"."
            print(ch1)
            print(ch2)
            print(ch3)
            print(ch1)
      
