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.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;


public class B {

	static StringTokenizer st;
	static BufferedReader br;
	static PrintWriter pw;
	public static void main(String[] args) throws IOException {
		br = new BufferedReader(new InputStreamReader(System.in));
		pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		int n = nextInt();
		int k = nextInt();
		char[][]a = new char[4][n];
		for (int i = 0; i < 4; i++) {
			Arrays.fill(a[i], '.');
		}
		if (k % 2==0) {
			for (int i = 1; i <= k / 2; i++) {
				a[1][i] = a[2][i] = '#';
			}
		}
		else{
			if (k <= 3) {
				if (k==1) {
					a[1][n/2] = '#';
				}
				else {
					a[1][n/2] = a[1][n/2-1] = a[1][n/2+1] = '#';
				}
			}
			else {
				a[1][1] = a[1][2] = a[1][3] = '#';
				a[2][1] = a[2][3] = '#';
				k -= 5;
				for (int i = 1; i <= k/2; i++) {
					a[1][i+3] = a[2][i+3] = '#';
				}
			}
		}
		System.out.println("YES");
		for (int i = 0; i < 4; i++) {
			for (int j = 0; j < n; j++) {
				System.out.print(a[i][j]);
			}
			System.out.println();
		}
		pw.close();
	}
	private static int nextInt() throws IOException {
		return Integer.parseInt(next());
	}
	private static long nextLong() throws IOException {
		return Long.parseLong(next());
	}
	private static double nextDouble() throws IOException {
		return Double.parseDouble(next());
	}
	private static String next() throws IOException {
		while (st==null || !st.hasMoreTokens())
			st = new StringTokenizer(br.readLine());
		return st.nextToken();
	}
}