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
.....
.###.
.....
.....
//Code by Sounak, IIEST
import java.io.*;
import java.math.*;
import java.util.*;
import java.util.Arrays;
public class Test1{
    public static void main(String args[])throws IOException{
        FastReader sc = new FastReader();
        StringBuilder sb = new StringBuilder();
        int n=sc.nextInt();
        int k=sc.nextInt();
        char a[][]=new char[4][n];
        int i,j=0;
        System.out.println("YES");
        for(i=0;i<4;i++)
        {
            for(j=0;j<n;j++)
            a[i][j]='.';
        }
        if(k%2==1)
        {
            a[1][n/2]='#';
            k--;
        }
        else if(k==2*(n-2))
        {
            a[2][n/2]='#';
            a[1][n/2]='#';
            k-=2;
        }
        for(i=1;i<n/2;i++)
        {
            if(k<=0)
            break;
            a[1][i]='#';
            a[1][n-1-i]='#';
            k-=2;
        }
        for(i=1;i<n/2;i++)
        {
            if(k<=0)
            break;
            a[2][i]='#';
            a[2][n-1-i]='#';
            k-=2;
        }
        if(k>0)
        a[2][n/2]='#';
        for(i=0;i<4;i++)
        {
            for(j=0;j<n;j++)
            System.out.print(a[i][j]);
            System.out.println();
        }
    }
    public static long pow(long x, long n)
    {
        long res = 1l; 
        while (n>0) 
        { 
            if ((n & 1)>0) 
            res = (res * x) % 1000000007l; 
            x = (x * x) % 1000000007l; 
            n >>= 1; 
        }
        return res;
    }
    public static long gcd(long a, long b) 
    { 
        if (a == 0) 
            return b; 
          
        return gcd(b%a, a); 
    }
}


class FastReader {
    BufferedReader br;
    StringTokenizer st;
 
    public FastReader() { br = new BufferedReader(new InputStreamReader(System.in)); }
    String next() {
        while (st == null || !st.hasMoreElements()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }
    int nextInt() { return Integer.parseInt(next()); }
    long nextLong() { return Long.parseLong(next()); }
    double nextDouble() { return Double.parseDouble(next()); }
    String nextLine() {
        String str = "";
        try {
            str = br.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return str;
    }
}