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.BufferedInputStream;
import java.io.BufferedOutputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.Scanner;


/*

.......
.#.....
.#.....
.......

.....
.###.
.....
.....
*/
public class Main{
    
    static HashSet<Integer> set2 = new HashSet<Integer>();
    static ArrayList<Integer> set = new ArrayList<Integer>();
    static char a[][] = new char[105][105];
    public static void main(String[] args){
        Scanner sc = new Scanner(new BufferedInputStream(System.in));
        PrintWriter pw = new PrintWriter(new BufferedOutputStream(System.out));
        
        int n,k;
        n = sc.nextInt();
        k = sc.nextInt();
        int nowrow=2,le=2,ri=n-1,time = n-2;
        int cnt=0;
        
        if(k%time==1||(k<time&&k%2==1)){
            a[3][n/2+1] = '#';
            k--;
        }
        
        while(k>0){
            k--;
            if(ri<le){
                nowrow++;
                le=2;
                ri = n-1;
            }
            if(cnt%2!=0) a[nowrow][ri--] = '#';
            else a[nowrow][le++] = '#';
            cnt++;
        }
        pw.println("YES");
        for(int i=1;i<=4;i++){
            for(int j=1;j<=n;j++)
            if(a[i][j]=='#') pw.print(a[i][j]);
            else pw.print('.');
            pw.println();
        }
        pw.flush();
    }
}