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.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;
public class Marlin {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		// TODO Auto-generated method stub

		MyScanner input = new MyScanner();
		int n = input.nextInt();
		int k = input.nextInt();
		char[][] s = new char[4][n];
		for(int i=0; i<4;i++)Arrays.fill(s[i],'.');
		if(k%2==0 && k>(n-2)*2){
			System.out.println("NO");
			return;
		}
		if(k%2==1 && k>(n-2)*2){
			System.out.println("NO");
			return;
		}
		System.out.println("YES");
		if(k==0){
			for(int i=0; i<4;i++){
				for(int j=0;j<n;j++)System.out.print(s[i][j]);
				System.out.println();
			}
			return;
		}
		if(k%2==0 && k>0){
			for(int i=1; i<n-1;i++){
				s[1][i]='#';
				s[2][i]='#';
				k-=2;
				if(k==0)break;
				
			}
			for(int i=0; i<4;i++){
				for(int j=0;j<n;j++)System.out.print(s[i][j]);
				System.out.println();
			}
			return;
			
		}
		else if(k%2==1){
			//int count=0;
		//	System.out.println(k);
			for(int i=1; i<3;i++){
				
					//System.out.println(s[i][n/2]);
				if(k%2!=0){
				s[i][n/2]='#';	
				k--;
					int l=n/2-1;
					int m=n/2+1;
					while(k>0 && l>0){
						s[i][l]='#';
					    s[i][m]='#';
						l--;
						m++;
						k-=2;
					}
					if(k==0)break;
				}
				else{
					int l = 1;
					int m=n-2;
					while(k>0){
						s[i][l]='#';
						s[i][m]='#';
						l++;
						m--;
						k-=2;
					  
					}
					if(k==0)break;
				}
			    
			}
			//System.out.println(2*(n-2));
			for(int  i=0; i<4;i++){
				for(int j=0; j<n;j++)System.out.print(s[i][j]);
				System.out.println();
			}
			return;
		}
	}

	



	static class MyScanner {
			BufferedReader br;
			StringTokenizer st;
			public MyScanner() {
				br = new BufferedReader(new InputStreamReader(System.in));
			}
			String next() {while (st == null || !st.hasMoreElements()) {
				try {st = new StringTokenizer(br.readLine());}
				catch (IOException e) {e.printStackTrace();}}
			return st.nextToken();}
			int nextInt() {return Integer.parseInt(next());}
			long nextLong() {return Long.parseLong(next());}
			double nextDouble() {return Double.parseDouble(next());}
			String nextLine(){String str = "";
			try {str = br.readLine();}
			catch (IOException e) {e.printStackTrace();}
			return str;}
		}
	
}
