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.Arrays;
import java.util.Scanner;

public class b{
    public static void main(String[] args){
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt();
        int k = scan.nextInt();
        char[][] mat = new char[4][n];
        if(k%2==0){
            for(int i=0;i<n;++i)mat[0][i]=mat[3][i]='.';
            mat[1][0]=mat[2][0]='.';
            for(int i=0;i<(k/2);++i)mat[1][i+1]=mat[2][i+1]='#';
            for(int i=1+k/2;i<n;++i)mat[1][i]=mat[2][i]='.';

            System.out.println("YES");
            for(int i=0;i<4;++i){
                for(int j=0;j<n;++j)System.out.print(mat[i][j]);
                System.out.println();
            }

        }else{
            if(k<n-2){
                for(char[] c : mat)Arrays.fill(c, '.');
                for(int i=n/2-k/2;i<=n/2+k/2;++i)mat[1][i]='#';
            }else{
                for(int i=0;i<n;++i)mat[0][i]=mat[3][i]=mat[2][i]='.';
                mat[1][0]=mat[2][0]=mat[1][n-1]=mat[2][n-1]='.';
                for(int i=0;i<n-2;++i)mat[1][i+1]='#';
                if(k>n-2){
                        mat[2][1]=mat[2][n-2]='#';
                        for(int i=0;i<k-n;++i)mat[2][2+i]='#';
                }
            }
            System.out.println("YES");
            for(int i=0;i<4;++i){
                for(int j=0;j<n;++j)System.out.print(mat[i][j]);
                System.out.println();
            }
        }
    }
}