660_B. Seating On Bus_5269
Consider 2n rows of the seats in a bus. n rows of the seats on the left and n rows of the seats on the right. Each row can be filled by two people. So the total capacity of the bus is 4n.

Consider that m (m ≤ 4n) people occupy the seats in the bus. The passengers entering the bus are numbered from 1 to m (in the order of their entering the bus). The pattern of the seat occupation is as below:

1-st row left window seat, 1-st row right window seat, 2-nd row left window seat, 2-nd row right window seat, ... , n-th row left window seat, n-th row right window seat.

After occupying all the window seats (for m > 2n) the non-window seats are occupied:

1-st row left non-window seat, 1-st row right non-window seat, ... , n-th row left non-window seat, n-th row right non-window seat.

All the passengers go to a single final destination. In the final destination, the passengers get off in the given order.

1-st row left non-window seat, 1-st row left window seat, 1-st row right non-window seat, 1-st row right window seat, ... , n-th row left non-window seat, n-th row left window seat, n-th row right non-window seat, n-th row right window seat.

<image> The seating for n = 9 and m = 36.

You are given the values n and m. Output m numbers from 1 to m, the order in which the passengers will get off the bus.

Input

The only line contains two integers, n and m (1 ≤ n ≤ 100, 1 ≤ m ≤ 4n) — the number of pairs of rows and the number of passengers.

Output

Print m distinct integers from 1 to m — the order in which the passengers will get off the bus.

Examples

Input

2 7


Output

5 1 6 2 7 3 4


Input

9 36


Output

19 1 20 2 21 3 22 4 23 5 24 6 25 7 26 8 27 9 28 10 29 11 30 12 31 13 32 14 33 15 34 16 35 17 36 18
import java.io.*;
import java.util.*;

public class B
{
    public static void main(String[] args)
    {
        MyScanner sc = new MyScanner();
        out = new PrintWriter(new BufferedOutputStream(System.out));

        int n = sc.nextInt(), m = sc.nextInt();
        	
        int cnt = 0, b = 1, bb = 2 * n + 1;
        for(int i = 0; i < n && cnt < m; i++) {
        	for(int j = 0; j < 4 && cnt < m; j++) {
        		if(j == 0 && bb <= m)
        			out.printf("%d ", bb);
        		else if(j == 1 && b <= m)
        			out.printf("%d ", b);
        		else if(j == 2 && bb + 1 <= m)
        			out.printf("%d ", bb + 1);
        		else if(j == 3 && b + 1 <= m)
        			out.printf("%d ", b + 1);
        	}
        	bb += 2;
        	b += 2;
        }
        
        out.close();
    }

    // PrintWriter for faster output
    public static PrintWriter out;

    // MyScanner class for faster input
    public static class MyScanner
    {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner()
        {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

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

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

        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;
        }
    }
}
