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.StringTokenizer;

public class Main {
	static boolean[][] field = new boolean[4][101];

	public static void main(String[] args) {
		// Scanner reader = new Scanner(System.in);
		InputReader reader = new InputReader();
		int n = reader.nextInt();
		int k = reader.nextInt();
		System.out.println("YES");
		for (int i = 1; i < n - 1; i++) {
			for (int j = 1; j < 3; j++) {
				if (k <= 1) {
					break;
				}
				k -= 2;
				field[j][i] = true;
				field[j][n - 1 - i] = true;
				if (n - 1 - i == i) {
					k++;
				}
			}
		}

		if (k > 0 && !field[1][n / 2]) {
			field[1][n / 2] = true;
			k--;
		}

		if (k > 0) {
			field[2][n / 2] = true;
		}

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

class InputReader {
	BufferedReader buf;
	StringTokenizer tok;

	InputReader() {
		buf = new BufferedReader(new InputStreamReader(System.in));
	}

	boolean hasNext() {
		while (tok == null || !tok.hasMoreElements()) {
			try {
				tok = new StringTokenizer(buf.readLine());
			} catch (Exception e) {
				return false;
			}
		}
		return true;
	}

	String next() {
		if (hasNext())
			return tok.nextToken();
		return null;
	}

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

	char nextChar() {
		return next().charAt(0);
	}

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

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

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

	static void deBug(int n, int m, int[][] array) {
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < m; j++) {
				System.out.print(array[i][j] + " ");
			}
			System.out.println();
		}
	}

	/*
	 * BigInteger nextBigInteger() { return new BigInteger(next()); }
	 * 
	 * BigDecimal nextBigDecimal() { return new BigDecimal(next()); }
	 */
}
// wow, thanks for your code review!