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.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.Scanner;

public class Second {
 
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner s=new Scanner(System.in);
		int n=s.nextInt();
		int k=s.nextInt();
		if(n<=(2*k+1)){
			System.out.println("1");
			if(n%2!=0){
			    System.out.println(n/2+1);
			}
			else{
			    System.out.println(n/2);    
			}
			
		}
		else{
			ArrayList<Integer> al=new ArrayList<Integer>();
			//leave k+1 skewers.
			int ans=0;
			int i=k+1;
			while(i+(2*k+1)<=n){
				ans++;
				al.add(i+k+1);
				i+=(2*k+1);
				
			}
			int rem=(n-i);
			int t=n-rem+k+1;
			int trem=rem;
			rem+=k+1;
			if(rem<=(2*k+1)){
				ans++;
				System.out.println(ans);
				al.add(1+trem);
				for(int q=0;q<al.size()-1;q++){
					al.set(q, al.get(q)+trem);
				}
				Collections.sort(al);
				for(int q=0;q<al.size();q++){
					System.out.print(al.get(q)+" ");
				}
			}
			else{
				ans+=2;
				System.out.println(ans);
				al.add(t);
				al.add(1);
				Collections.sort(al);
				for(int q=0;q<al.size();q++){
					System.out.print(al.get(q)+" ");
				}
			}
			
		}
		
    }
}