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 Main2 {

    private FastScanner scanner = new FastScanner();

    public static void main(String[] args) {
        new Main2().solve();
    }

    private void solve() {

        int n = scanner.nextInt(), k = scanner.nextInt();

        char[][] ans = new char[4][n];
        for (char[] ch : ans) {
            Arrays.fill(ch, '.');
        }

        if (k % 2 == 0) {
            for (int i = 0; i < k / 2; i++) {
                ans[1][i + 1] = '#';
                ans[2][i + 1] = '#';
            }
        } else {

            ans[1][n / 2] = '#';
            for (int i = 0; i < Math.min((n - 3) / 2, (k - 1) / 2); i++) {
                ans[1][n / 2 + (i + 1)] = '#';
                ans[1][n / 2 - (i + 1)] = '#';
            }

            if (k > n - 2) {
                ans[2][n - 2] = '#';
                for (int i = 0; i < k - (n - 2) - 1; i++) {
                    ans[2][i + 1] = '#';
                }
            }

        }

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

    }

    long gcd(long a, long b) {
        if (b != 0) {
            return gcd(b, a % b);
        }
        return a;
    }

    class FastScanner {
        BufferedReader reader;
        StringTokenizer tokenizer;

        FastScanner() {
            reader = new BufferedReader(new InputStreamReader(System.in), 32768);
            tokenizer = null;
        }

        String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();

        }

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

        int[] nextA(int n) {
            int a[] = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = scanner.nextInt();
            }
            return a;
        }

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

        public double nextDouble() {
            return Double.parseDouble(next());
        }

        public String nextLine() {
            try {
                return reader.readLine();
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
    }
}