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.io.*;

public class Main
{
	static long mod=1000000007;
	public static void main(String[] args) throws IOException
	{
		FastReader in=new FastReader();
		PrintWriter pw=new PrintWriter(System.out);
		int n=in.nextInt();
		int k=in.nextInt();
		if(k==0)
		{
			System.out.println(n);
			for(int i=1;i<=n;i++)
				pw.print(i+" ");
			pw.flush();
			return;
		}
		int ans=Integer.MAX_VALUE,ind=0;
		for(int i=1;i<=k+1;i++)
		{
			ind=i;
			int j=i;
			int t=1;
			while(j+2*k+1<=n)
			{
				j=j+2*k+1;
				t++;
			}
			if(j+k+1<=n)
				continue;
			ans=t;
			break;
		}
		System.out.println(ans);
		int i=ind;
		while(i<=n)
		{
			pw.print(i+" ");
			i=i+2*k+1;
		}
		pw.flush();
	}
}

class FastReader
{
	BufferedReader br;
	StringTokenizer st;
 
	public FastReader()
	{
		br=new BufferedReader(new InputStreamReader(System.in));
	}
 
	public String next() throws IOException
	{
		if(st==null || !st.hasMoreElements())
		{
			st=new StringTokenizer(br.readLine());
		}
		return st.nextToken();
	}
 
	public int nextInt() throws IOException
	{
		return Integer.parseInt(next());
	}
 
	public long nextLong() throws IOException
	{
		return Long.parseLong(next());
	}
}