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.Scanner;

public class B {
    static boolean[][] solveEven(int n, int k) {
        boolean[][] ret = new boolean[4][n];
        for (int l = 0; l < k / 2; ++l) {
            ret[1][l + 1] = true;
            ret[2][l + 1] = true;
        }
        return ret;
    }

    static boolean[][] solveOdd(int n, int k) {
        boolean[][] ret = new boolean[4][n];
        int mid = n / 2;
        ret[1][mid] = true;
        --k;
        int step = 1;
        while (k > 0 && step <= mid - 1) {
            ret[1][mid + step] = true;
            ret[1][mid - step] = true;
            ++step;
            k -= 2;
        }

        step = 1;
        while (k > 0) {
            ret[2][step] = true;
            ret[2][n - 1 - step] = true;
            k -= 2;
            ++step;
        }

        return ret;
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int k = scanner.nextInt();
        boolean[][] answer;
        if (k % 2 == 0) {
            answer = solveEven(n, k);
        } else {
            answer = solveOdd(n, k);
        }

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