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
.....
.###.
.....
.....
#!/usr/bin/python2

#t=int(raw_input())
#for i in xrange(t):
n,k=map(int,raw_input().split())
d={}
for i in xrange(2):
    i=i+1
    for j in xrange(n-2):
        j=j+1
        d[(i,j)]=1

if k>2*(n-2):
    print 'NO'
else:
    print 'YES'
    while((k!=1) and (k!=0) and len(d.keys())>2):
        key=d.keys()[0]
        if key[1]==(n-1)/2:
            key=d.keys()[1]
        if key[1]==(n-1)/2:
            key=d.keys()[2]
        del d[key]
        key=(key[0],n-1-key[1])
        del d[key]
        k-=2

    if k==1:
        del d[(1,(n-1)/2)]
    elif k==2:
        del d[d.keys()[0]]
        del d[d.keys()[0]]

    for i in xrange(4):
        s=''
        for j in xrange(n):
            if i==0 or i==3 or j==0 or j==n-1:
                s+='.'
            elif d.has_key((i,j)):
                s+='.'
            else:
                s+='#'
        print s
