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.*;
import java.util.*;

public class R480B {

	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		int r = 4;
		int c = scan.nextInt();
		int k = scan.nextInt();
		int area = (r-2)*(c-2);
		if(k % 2 == 1){
			if(c % 2 == 0){
				out.println("NO");
				out.flush();
				return;
			}
		}
		if(area >= k){
			if(k % 2 == 0){
				out.println("YES");
				
				for(int i = 0; i < c; i++) out.print('.');
				out.println();
				
				int count = 1;
				out.print('.');
				for(int i = 0; i < k/2; i++){
					out.print('#');
					count++;
				}
				while(count < c){
					out.print('.');
					count++;
				}
				out.println();
				
				
				count = 1;
				out.print('.');
				for(int i = 0; i < k/2; i++){
					out.print('#');
					count++;
				}
				while(count < c){
					out.print('.');
					count++;
				}
				out.println();
				
				for(int i = 0; i < c; i++) out.print('.');
				out.println();
				
			}
			else if(k % 2 == 1){
				
				out.println("YES");
				
				for(int i = 0; i < c; i++) out.print('.');
				out.println();
				
				char[] arr1 = new char[c];
				char[] arr2 = new char[c];
				Arrays.fill(arr1, '.');
				Arrays.fill(arr2, '.');
				int idx = c/2;
				int bonus = 1;
				arr1[idx] = '#';
				k--;
				
				while(idx-bonus > 0 && k > 0){
					arr1[idx+bonus] = '#';
					arr1[idx-bonus] = '#';
					k-=2;
					bonus++;
				}
				
				bonus--;
				while(k > 0){
					arr2[idx+bonus] = '#';
					arr2[idx-bonus] = '#';
					k-=2;
					bonus--;
				}
				
				for(int i = 0; i < c; i++) out.print(arr1[i]);
				out.println();
				for(int i = 0; i < c; i++) out.print(arr2[i]);
				out.println();
				
				
				for(int i = 0; i < c; i++) out.print('.');
				out.println();
			}
			else out.println("NO");
		}
		else{
			out.println("NO");
		}
		out.flush();
	}
}
