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.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner s = new Scanner(System.in);
		int n = s.nextInt(), k = s.nextInt();
		char[][] arr = new char[4][n];
		for(int i = 0; i < arr.length; i++){
			for(int j = 0; j < arr[0].length; j++){
				arr[i][j] = '.';
			}
		}
	/*	if(k == 2 * (n - 2)){
			System.out.println("NO");
			return;
		}*/
		if(k % 2 == 1){
			int center = n / 2;
			if(k > n - 2){
				int def = 1;
				for(int i = 1; i < n - 1; i++){
					arr[1][i] = '#';
				}
				k -= n - 2;
				int i = 1;
			    while(k > 0){
			    	arr[2][i] = arr[2][n - i - 1] = '#';
			    	k -= 2;
			    	i++;
			    }
			}else{
				int def = 1;
				arr[1][center] = '#';
				k--;
				while(k > 0 && center - def > 0){
					arr[1][center - def] = '#';
					k--;
					arr[1][center + def] = '#';
					k--;
					def++;
				}
			}
		}else{
			for(int i = 1; i <= k / 2; i++){
				arr[1][i] = arr[2][i] = '#';
			}
		}
		System.out.println("YES");
		for(int i = 0; i < 4; i++){
			for(int j = 0; j < n; j++){
				System.out.print(arr[i][j]);
			}
			System.out.println();
		}

	}

}
