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.*;

public class Main {
	
	static void debug(Object... O) {
		System.out.println(Arrays.deepToString(O));
	}

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int k = sc.nextInt();
		StringBuilder sb = new StringBuilder();
		char[][] F = new char[4][n];
		for(int i=0;i<4;i++) {
			Arrays.fill(F[i], '.');
		}
		boolean ans = true;
		int mid = n/2;
		int left = mid-1,right=mid+1;
		boolean first = true, second = true;
		while(k > 0) {
			if(first && left > 0 &&  k >= 2) {
				
				F[1][left] = '#';
				F[1][right] = '#';
				left--;
				right++;
				k-=2;
				//System.out.println("up k "+k);
				if(left == 0) {
					first = false;
					left = mid-1;
					right=mid+1;
				}
			}else if(!first && second && left > 0 && k>=2) {
				F[2][left] = '#';
				F[2][right] = '#';
				left--;
				right++;
				k-=2;
				//System.out.println("down k "+k);
				if(left == 0 && k >2) {
					ans = false;
					break;
				}else if(left == 0) {
					second = false;
				}
			}else if(k == 1) {
				F[1][mid] ='#';
				k--;
			}else if(k == 2) {
				F[1][mid] = '#';
				F[2][mid] = '#';
				k-=2;
			}
			
			
		}
		if(ans) {
			System.out.println("YES");
			for(int i=0;i<4;i++) {
				for(int j=0;j<n;j++) {
					sb.append(F[i][j]);
				}
				sb.append("\n");
			}
			System.out.println(sb.toString());
		}else {
			System.out.println("NO");
		}
		
	}

}
