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.*;
import java.lang.*;

public class Shasha{
	public static void main(String args[])throws IOException{
		Scanner ob=new Scanner(System.in);
		//BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		int n=ob.nextInt();
		int k=ob.nextInt();
		int res;
		int x=2*k+1;
		if(n%(2*k+1)==0){
			res=n/(2*k+1);
			System.out.println(res);
			for(int i=1+k;i<=n;i+=(2*k+1))
				System.out.print(i+" ");
			System.out.println();
		}
		else{
			if(n%x<=k){
				res=(n/x)+1;
				System.out.println(res);
				int l[]=new int[2];
				for(int i=k+1;i<=2*k;i++){
					for(int j=k+1;j<=2*k;j++){
						if((i+j)==(n%x+x)){
							l[0]=i;
							l[1]=j;
							break;
						}
					}
				}
				//System.out.println(l[0]+" "+l[1]);
				System.out.print((l[0]-k)+" ");
			for(int i=l[0]-k+x;i<=n;i+=(x))
				System.out.print(i+" ");		
			//System.out.print((n-(k+1-l[1]))+" ");
			System.out.println();
			}
			else{
			res=(n/x)+1;
			System.out.println(res);
			System.out.print((n%x-k)+" ");
			for(int i=n%x-k+x;i<=n;i+=(x))
				System.out.print(i+" ");
			System.out.println();
			}
		}
	}
}