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.Arrays;
import java.util.Scanner;

//accidentally stabbed my hand while cutting avacado :(
//still programming tho :)

public class Marlin {
	public static void main(String[] args) {
		Scanner scan=new Scanner(System.in);
		int n=scan.nextInt(), k=scan.nextInt();
		char[][] a=new char[4][n];
		for(char[] c:a)Arrays.fill(c,'.');
		if(k%2==0) {
			for(int i=0;i<k/2;i++) {
				a[1][i+1]=a[2][i+1]='#';
			}
		}
		else if(k>n-2) {
			a[1][1]=a[2][1]=a[1][n-2]=a[2][n-2]='#';
			k-=(n==3)?2:4;
			
			//fill in the rest
			for(int i=1;i<3;i++) {
				for(int j=2;j<n-2;j++) {
					if(k>0) {
						a[i][j]='#';
						k--;
					}
				}
			}
		}
		else {
			for(int i=1;i<3;i++) {
				for(int j=0;j<n-1;j++) {
					if(k>0) {
						a[i][(int)((n+0.5)/2)-j]='#';
						a[i][(int)((n+0.5)/2)+j]='#';
						k-=j==0?1:2;
					}
				}
			}
		}
		System.out.println("YES");
		for(char[] c:a) System.out.println(new String(c));
	}
}