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.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;

/**
 * Created by omar on 07/04/18.
 */
public class B {

    public static void main(String[] args) throws Exception {
        Scanner sc = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);

        int n = sc.nextInt(), k = sc.nextInt();
        out.println("YES");
        char[][] a = new char[4][n];
        for (char[] aa:a) Arrays.fill(aa,'.');
        for (int c = 0; c < (n - 2) / 2; c++) {
            for (int row = 1; row < 3; row++) {
                if(k>1){
                    a[row][c+1]='#';
                    a[row][n-2-c]='#';
                    k-=2;
                }
            }
        }
        int row=1;
        while(k>0){
            k--;
            a[row++][n/2]='#';
        }

        for (char[] aa : a)
            out.println(new String(aa));

        out.flush();
    }


    static class Scanner {
        StringTokenizer st;
        BufferedReader bf;

        Scanner(InputStream in) {
            bf = new BufferedReader(new InputStreamReader(in));
        }

        String next() throws Exception {
            while (st == null || !st.hasMoreTokens()) {
                st = new StringTokenizer(bf.readLine());
            }
            return st.nextToken();
        }

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

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