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.*;
import java.util.*;


public class test {
    int INF = (int)1e9;
    long MOD = 1000000007;
    
    void go(int[][] a, int i, int k, int n) {
        int y = (n-k)/2;
        int j = y;
        while(k-->0) {
            a[i][j] = 1;
            j++;
        }
    }
    
    void solve(InputReader in, PrintWriter out) throws IOException {
        int n = in.nextInt();
        int k = in.nextInt();
        int[][] a = new int[4][n];
        if(k%2==0) {
            for(int i=1; i<=k/2; i++) {
                a[1][i] = a[2][i] = 1;
            }
        } else {
            a[1][n/2] = 1;
            k--;
            for(int i=1; i<n/2; i++) {
                if(k<=0) break;
                a[1][i] = 1;
                a[1][n-1-i] = 1;
                k-=2;
                if(k<=0) break;
            }
            for(int i=1; i<n/2; i++) {
                if(k<=0) break;
                a[2][i] = 1;
                a[2][n-1-i] = 1;
                k-=2;
                if(k<=0) break;
            }
        }
        out.println("YES");
        for(int i=0; i<4; i++) {
            for(int j=0; j<n; j++) {
                if(a[i][j]==0) out.print('.');
                else out.print('#');
            }
            out.println();
        }
    }
    
    
    public static void main(String[] args) throws IOException {
        if(args.length>0 && args[0].equalsIgnoreCase("d")) {
            DEBUG_FLAG = true;
        }
        InputReader in = new InputReader();
        PrintWriter out = new PrintWriter(System.out);
        int t = 1;//in.nextInt();
        long start = System.nanoTime();
        while(t-->0) {
            new test().solve(in, out);
        }
        long end = System.nanoTime();
        debug("\nTime: " + (end-start)/1e6 + " \n\n");
        out.close();
    }
    
    static boolean DEBUG_FLAG = false;
    static void debug(String s) {
        if(DEBUG_FLAG) System.out.print(s);
    }
    
    public static void show(Object... o) {
        System.out.println(Arrays.deepToString(o));
    }
    
    static class InputReader {
        static BufferedReader br;
        static StringTokenizer st;
    
        public InputReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }
        
        String next() {
            while (st == null || !st.hasMoreTokens()) {
                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());
        }
    }
}