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
.....
.###.
.....
.....
import java.util.*;
import java.io.*;

public class Marlin 
{
	public static void main(String[] args)
	{
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int k = in.nextInt();
		
		int[][] ans = new int[4][n];
		
		if(k%2==0)
		{
				System.out.println("YES");
				
				for(int i=0;i<2;i++)
				{
					for(int j=0;j<k/2;j++)
						ans[i+1][j+1] = '#';
				}
				
				for(int i=0;i<4;i++)
				{
					for(int j=0;j<n;j++)
					{
						if(ans[i][j]=='#')
							System.out.print("#");
						else
							System.out.print(".");
					}
					System.out.println("");
				}
		}
		else
		{
			if(k<=n-2)
			{
				if(n%2!=0)
				{
					System.out.println("YES");
					
					ans[1][n/2]='#';
					
					for(int i=n/2;i<n/2+k/2;i++)
					{
						ans[1][n-i-2]='#';
						ans[1][i+1]='#';
					}
					
					for(int i=0;i<4;i++)
					{
						for(int j=0;j<n;j++)
						{
							if(ans[i][j]=='#')
								System.out.print("#");
							else
								System.out.print(".");
						}
						System.out.println("");
					}
				}
				else
				{
					System.out.println("NO");
				}
			}
			else
			{
				k = k - n + 2;

				if(k==1)
					System.out.println("NO");
				else
				{
					System.out.println("YES");
					
					for(int i=1;i<n-1;i++)
						ans[1][i] = '#';
					
					ans[2][n-2] = '#';
					
					for(int i=1;i<k;i++)
					{
						ans[2][i] = '#';
					}
					
					for(int i=0;i<4;i++)
					{
						for(int j=0;j<n;j++)
						{
							if(ans[i][j]=='#')
								System.out.print("#");
							else
								System.out.print(".");
						}
						System.out.println("");
					}
				}
			}
		}
	}
}
