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(i) for i in input().strip().split()]
l=[]
for i in range(4):
    l.append([])
    for j in range(n):
        l[-1].append('.')
if (n==3 or k%2==0):
    print ('YES')
    x=k//2
    for i in range(1,x+1):
        for j in range(1,3):
            l[j][i]='#'
    if (k%2!=0):
        l[1][1]='#'
    for i in range(4):
        for j in range(n):
            print (l[i][j],end='')
        print (end='\n')
else:
    if (n==4):
        print ('NO')
    else:
        if(n%2==0):
            if (k!=(2*n-5)):
                print ('NO')
            else:
                print ('YES')
                x=k//2 - 1
                for i in range(1,x+1):
                    for j in range(1,3):
                        l[j][i]='#'
                l[1][n-3]='#'
                l[1][n-2]='#'
                l[2][n-2]='#'
                for i in range(4):
                    for j in range(n):
                        print (l[i][j],end='')
                    print (end='\n')
        else:
            print ('YES')
            l[1][n//2]='#'
            if (k-1<=n-3):
                count=k-1
                a=n//2 -1
                b=n//2 +1
                while (count!=0):
                    l[1][a]='#'
                    l[1][b]='#'
                    a-=1
                    b+=1
                    count-=2
            else:
                for i in range(1,n-1):
                    l[1][i]='#'
                count=k-(n-2)
                a=1
                b=n-2
                while (count!=0):
                    l[2][a]='#'
                    l[2][b]='#'
                    a+=1
                    b-=1
                    count-=2
            for i in range(4):
                for j in range(n):
                    print (l[i][j],end='')
                print (end='\n')