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
.....
.###.
.....
.....
// practice with kaiboy
import java.io.*;
import java.util.*;

public class CF980B extends PrintWriter {
	CF980B() { super(System.out); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF980B o = new CF980B(); o.main(); o.flush();
	}

	void main() {
		int n = sc.nextInt();
		int k = sc.nextInt();
		char[][] cc = new char[4][n];
		for (int i = 0; i < 4; i++)
			for (int j = 0; j < n; j++)
				cc[i][j] = '.';
		int l = 1, r = n - 2;
		while (k >= 4) {
			cc[1][l] = cc[2][l] = cc[1][r] = cc[2][r] = '#';
			l++; r--;
			k -= 4;
		}
		if (k >= 2) {
			if (l == r)
				cc[1][l] = cc[2][l] = '#';
			else
				cc[1][l] = cc[1][r] = '#';
			k -= 2;
		}
		if (k == 1)
			cc[1][(l + r) / 2] = '#';
		println("YES");
		for (int i = 0; i < 4; i++)
			println(cc[i]);
	}
}
