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
.....
.###.
.....
.....
z,zz=input,lambda:list(map(int,z().split()));
szz,dgraphs,mod=lambda:sorted(zz()),{},10**9+7
from string import *
from collections import *
from queue import *
from sys import *
from collections import *
from math import *
from heapq import *
from itertools import *
from bisect import *
from collections import Counter as cc
from math import factorial as f
from bisect import bisect as bs
from itertools import accumulate as ac
def lcd(xnum1,xnum2):return (xnum1*xnum2//gcd(xnum1,xnum2))
def prime(x):
    p=ceil(x**.5)+1
    for i in range(2,p):
        if x%i==0 and x!=2:return 0
    return 1
def dfs(u,visit,graph):
    visit[u]=True
    for i in graph[u]:
        if not visit[i]:
            dfs(i,visit,graph)
###########################---Test-Case---#################################
"""




"""
###########################---START-CODING---##############################

n, k =zz()
a = '.' * n

print('YES')
print(a)
if k % 2 == 0:
    k //= 2
    print('.' + '#'*k + '.'*(n-k-1))
    print('.' + '#'*k + '.'*(n-k-1))
elif k <= n-2:
    print('.' * ((n - k) // 2) + '#' * k + '.' * ((n - k) // 2))
    print(a)
else:
    k -= n-2
    print('.' + '#'*(n-2) + '.')
    print('.' + '#'*(k//2) + '.'*(n-2-k) +  '#'*(k//2) + '.')

print(a)

        

