Problem: Long story short, shashlik is Miroslav's favorite food. Shashlik is prepared on several skewers simultaneously. There are two states for each skewer: initial and turned over.

This time Miroslav laid out n skewers parallel to each other, and enumerated them with consecutive integers from 1 to n in order from left to right. For better cooking, he puts them quite close to each other, so when he turns skewer number i, it leads to turning k closest skewers from each side of the skewer i, that is, skewers number i - k, i - k + 1, ..., i - 1, i + 1, ..., i + k - 1, i + k (if they exist). 

For example, let n = 6 and k = 1. When Miroslav turns skewer number 3, then skewers with numbers 2, 3, and 4 will come up turned over. If after that he turns skewer number 1, then skewers number 1, 3, and 4 will be turned over, while skewer number 2 will be in the initial position (because it is turned again).

As we said before, the art of cooking requires perfect timing, so Miroslav wants to turn over all n skewers with the minimal possible number of actions. For example, for the above example n = 6 and k = 1, two turnings are sufficient: he can turn over skewers number 2 and 5.

Help Miroslav turn over all n skewers.

Input

The first line contains two integers n and k (1 ≤ n ≤ 1000, 0 ≤ k ≤ 1000) — the number of skewers and the number of skewers from each side that are turned in one step.

Output

The first line should contain integer l — the minimum number of actions needed by Miroslav to turn over all n skewers. After than print l integers from 1 to n denoting the number of the skewer that is to be turned over at the corresponding step.

Examples

Input

7 2


Output

2
1 6 


Input

5 1


Output

2
1 4 

Note

In the first example the first operation turns over skewers 1, 2 and 3, the second operation turns over skewers 4, 5, 6 and 7.

In the second example it is also correct to turn over skewers 2 and 5, but turning skewers 2 and 4, or 1 and 5 are incorrect solutions because the skewer 3 is in the initial state after these operations.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.StringTokenizer;

/**
 * Created by User on 11.09.2018.
 */
public class Kek {
    static class Reader {
        BufferedReader in;

        Reader(String name) throws FileNotFoundException {
            in = new BufferedReader(new FileReader(name));
        }

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

        StringTokenizer st = new StringTokenizer("");

        void update() throws IOException {
            if (!st.hasMoreTokens()) {
                String a = in.readLine();
                if (a != null) {
                    st = new StringTokenizer(a);
                }
            }
        }

        String next() throws IOException {
            update();
            return st.nextToken();
        }

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

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

    }

    public static String fileName;
    public static Reader in;
    public static PrintWriter out;


    static int calc(int c, int n){
        return (int) (c*Math.ceil(1.0*n/(5-c%5)));
    }

    public static void main(String args[]) throws IOException {
        in=new Reader();
        //out=new PrintWriter("f.out");
        int n=in.nextInt();
        int kk=in.nextInt();
        int k=2*kk+1;
        int x=n%k;
        if (x==0){
            int st=kk+1;
            System.out.println(n/k);
            for(int i=0;i<n/k;i++){
                System.out.print(st+" ");
                st+=k;
            }
        }
        else if (x>=kk+1){
            int st=kk+1;
            System.out.println(n/k+1);
            for(int i=0;i<n/k+1;i++){
                System.out.print(st+" ");
                st+=k;
            }
        }
        else {
            int st=1;
            System.out.println(n/k+1);
            for(int i=0;i<n/k+1;i++){
                System.out.print(st+" ");
                st+=k;
            }
        }

        //out.close();
    }

}