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 CF980B {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int nCols = sc.nextInt();
		int k = sc.nextInt();
		char[][] grid = new char[4][nCols];
		for(int i=0; i<4; i++)
			for(int j=0; j<nCols; j++) grid[i][j] = '.';
		if(k%2==0) {
			System.out.println("YES");
			for(int i=1; i<=k/2; i++) grid[1][i] = grid[2][i] = '#';
			for(int i=0; i<4; i++) System.out.println(grid[i]);
		} else {
			if(k>nCols-2) {
				for(int i=1; i<nCols-1; i++) grid[1][i] = '#';
				k-= (nCols-2);
				for(int i=0; i<k/2; i++)
					grid[2][i+1] = grid[2][nCols-2-i] = '#';
				System.out.println("YES");
				for(int i=0; i<4; i++) System.out.println(grid[i]);
			} else {
				boolean can = false;
				int l = 0;
				for(int i=1; i<=nCols-2; i+=2)
					if(i==k) {
						can = true;
						l = i;
					}
				if(can) {
					int startCol = nCols/2-(l/2);
					int endCol = nCols/2+(l/2);
					int r = 1;
					for(int i=0; i<k; i+=l) {
						for(int j=startCol; j<=endCol; j++) grid[r][j] = '#';
						r++;
					}
					System.out.println("YES");
					for(int i=0; i<4; i++) System.out.println(grid[i]);
				} else System.out.println("NO");
			}
		}
	}
}
