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.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.StringTokenizer;
import java.util.TreeMap;

public class zizo {
	public static void main(String[]args) throws IOException {
		Scanner zizo=new Scanner(System.in);
		PrintWriter wr=new PrintWriter(System.out);
		
		int n=zizo.nextInt();
		int k=zizo.nextInt();
		wr.println("YES");//Always yes ha ha ha
		
		char[][]arr=new char[4][n];
		for(int i=0;i<4;i++) {
			Arrays.fill(arr[i],'.');
		}
		for(int i=1;i<3;i++) {
			for(int j=1;j<n/2 && k>1;j++) {
				arr[i][j]='#';
				arr[i][n-j-1]='#';
				k-=2;
			}
		}
//		for(int i=0;i<4;i++) {
//			for(int j=0;j<n;j++) {
//				wr.print(arr[i][j]+" ");
//			}
//			wr.println();
//		}
		for(int i=1;i<3 && k>0;i++) {
			arr[i][n/2]='#';
			k--;
		}
		for(int i=0;i<4;i++) {
			for(int j=0;j<n;j++) {
				wr.print(arr[i][j]);
			}
			wr.println();
		}
		wr.close();
	}
}

class Scanner 
{
	StringTokenizer st;
	BufferedReader br;

	public Scanner(InputStream s){	br = new BufferedReader(new InputStreamReader(s));}

	public String next() throws IOException 
	{
		while (st == null || !st.hasMoreTokens()) 
			st = new StringTokenizer(br.readLine(), ",| ");
		return st.nextToken();
	}

	public int nextInt() throws IOException {return Integer.parseInt(next());}

	public long nextLong() throws IOException {return Long.parseLong(next());}

	public String nextLine() throws IOException {return br.readLine();}

	public boolean ready() throws IOException {return br.ready();}

	public double nextDouble() throws IOException 
	{
		String x = next();
		StringBuilder sb = new StringBuilder("0");
		double res = 0, f = 1;
		boolean dec = false, neg = false;
		int start = 0;
		if(x.charAt(0) == '-')
		{
			neg = true;
			start++;
		}
		for(int i = start; i < x.length(); i++)
			if(x.charAt(i) == '.')
			{
				res = Long.parseLong(sb.toString());
				sb = new StringBuilder("0");
				dec = true;
			}
			else
			{
				sb.append(x.charAt(i));
				if(dec)
					f *= 10;
			}
		res += Long.parseLong(sb.toString()) / f;
		return res * (neg?-1:1);
	}


}