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 CF480_B {
	public static void main(String[] args) {
		Scanner s = new Scanner(System.in);
		int n = s.nextInt(), k = s.nextInt();
		StringBuilder ans = new StringBuilder();
		char[][] a = new char[4][n];
		int r = 1, c = n - 2, c1 = 1, cnt = 0;
		if (k == 2 * (n - 2)) {
			for (int i = 0; i < 4; i++) {
				for (int j = 0; j < n; j++) {
					if (i == 0 || j == 0 || i == 3 || j == n - 1)
						ans.append('.');
					else
						ans.append('#');
				}
				ans.append('\n');
			}
			System.out.println("YES\n" + ans);
			return;
		}
		for (int i = 0; i < k / 2; i++) {
			a[r][c] = '#';
			a[r][c1] = '#';
			c1++;
			c--;
			if (c1 == c) {
				r++;
				c1 = 1;
				c = n - 2;
			}
		}
		if (k % 2 == 1) {
			a[1][n / 2] = '#';
		}
		for (int i = 0; i < 4; i++) {
			for (int j = 0; j < n; j++) {
				if (a[i][j] == '#')
					ans.append(a[i][j]);
				else
					ans.append('.');
			}
			ans.append('\n');
		}
		System.out.println("YES\n" + ans);

	}

}
