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())))
print("YES")
a=["."]*n
b=[]
if k<=n-2:
    if k%2==0:
        for i in range(k//2):
            a[1+i]="#"
            a[n-1-1-i]="#"
    else:
        for i in range(k//2):
            a[1+i]="#"
            a[n-1-1-i]="#"
        a[n//2]="#"
    
    c=["."]*n
    b.append(c)
    b.append(a)
    b.append(c)
    b.append(c)
else:
    k-=(n-2)
    if k%2==0:
        for i in range(k//2):
            a[1+i]="#"
            a[n-1-1-i]="#"
    else:
        for i in range(k//2):
            a[1+i]="#"
            a[n-1-1-i]="#"
        a[n//2]="#"
    c=["."]*n
    b.append(c)
    e=["."]*n
    for i in range(1,n-1):
        e[i]="#"
    b.append(e)
    
    b.append(a)
    b.append(c)
for ele in b:
    print(*ele,sep="")