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.util.*;
import java.math.*;


public class q_2 {
    static Scanner sc;

    public static void main(String[] args){
        sc = new Scanner(System.in);
        int n = getInt();
        int k = getInt();
        if(n<=(2*k+1)){
            //System.out.println("a");
            System.out.println(1);
            System.out.println((n+1)/2);
        }
        else {
            //System.out.println("b");
            int rem = n%(2*k+1);
            if(rem==0){
                //System.out.println("c");
                System.out.println(n/(2*k+1));
                for(int i=k+1 ; i<=n ; i+=(2*k+1)){
                    System.out.print(i+" ");
                }
                return;
            }
            else {
                // now we have a solution
                //System.out.println("d");
                int c = rem+(2*k+1);
                System.out.println((n/(2*k+1))+1);
                for(int i=((c/2)-k);i<=n ; i+=(2*k+1)){
                    System.out.print(i+" ");
                }
            }
        }

    }

    public static int getInt(){
        return sc.nextInt();
    }

    public static String getString(){
        return sc.next();
    }

    public static String nextLine(){
        return sc.nextLine();
    }

}