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
.....
.###.
.....
.....
/* package whatever; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
public class Main
{
	public static void main (String[] args) throws java.lang.Exception
	{
		Scanner scan=new Scanner(System.in);
		int n=scan.nextInt();
		int k=scan.nextInt();
		String row1="";
		String row2="";
		String row3="";
		String row4="";
		for(int i=0;i<n;i++){
			row1+=".";
			row4+=".";
		}
		if(k%2==1){
			int[] A=new int[n];
			int[] B=new int[n];
			int r=k;
			for(int i=0;i<n;i++){
				A[i]=B[i]='.';
			}
			int p=1;
			int q=n-2;
			while(p<q && r>1){
				A[p]='#';
				A[q]='#';
				p++;
				q--;
				r-=2;
			}
			while(p<q){
				p++;
				q--;
			}
			A[p]='#';
			r--;
			p=1;
			q=n-2;
			while(p<q && r>0){
				B[p]='#';
				B[q]='#';
				p++;
				q--;
				r-=2;
			}
			for(int i=0;i<n;i++){
				row2+=(char)A[i];
				row3+=(char)B[i];
			}
			System.out.println("YES");
			System.out.println(row1);
			System.out.println(row2);
			System.out.println(row3);
			System.out.println(row4);
		}
		else{
			int r=k/2;
			row2+=".";
			row3+=".";
			for(int i=0;i<r;i++){
				row2+='#';
				row3+='#';
			}
			for(int i=r+1;i<n;i++){
				row2+='.';
				row3+='.';
			}
			System.out.println("YES");
			System.out.println(row1);
			System.out.println(row2);
			System.out.println(row3);
			System.out.println(row4);
		}
		
	}
}