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:
/* package whatever; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
public class Solution
{
	public static void main (String[] args) throws java.lang.Exception
	{
		// your code goes here
		Scanner scan=new Scanner(System.in);
		int n=scan.nextInt();
		int k=scan.nextInt();
		//int bi=0;
		if(k==0) {
			System.out.println(n);
			for(int i=1;i<=n;i++) {
				System.out.print(i + " ");
			}
			return;
		}
		int min=Integer.MAX_VALUE;
		int count=0;
		ArrayList<Integer> ans=new ArrayList<Integer>();
		for(int i=1;i<=k+1;i++){
			int next=i;
			count=1;
			ArrayList<Integer> current=new ArrayList<Integer>();
			current.add(next);
			while(true){
				if(next+1+2*k>n){
					break;
				}
				count++;
				next=next+1+2*k;
				current.add(next);
			}
			if(next+k>=n){
				if(count<min){
					ans=(ArrayList<Integer>)current.clone();
					min=count;
				}
			}
		}
		System.out.println(min);
		for(int i=0;i<ans.size();i++){
			System.out.print(ans.get(i) + " ");
		}
	}
}