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 int i,j,k,l,temp,count;
  static LinkedList<Integer> adj[];
	static boolean visited[];
	static Iterator<Integer> itr;
	public static void main(String[] args) throws IOException{
		/*
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		*/
		int n=ni();
		int k=ni();
		int count=n/((2*k)+1);
		int remainder=(n)%((2*k)+1);
		int constant=0;
		if(remainder!=0){
			count++;
			if(remainder<(k+1))
				constant =(k+1)-remainder;
		}
		sop(count);
		i=0;
		while(count>0){
			count--;
			System.out.print((i)*(2*k+1)+k+1-constant+" ");
			i++;
		}
		/*
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		.
		.
		.
		.
		.
		.
		*/
	}
	/*
	temporary functions
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	.
	*/

	/*
	fuctions
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	.
	.
	.
	.
	.
	.
	.
	.
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	abcdefghijklmnopqrstuvwxyz
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	*/
	static int modulo(int j,int m){
		if(j<0)
			return m+j;
		  if(j>=m)
			return j-m;

			return j;
	}
	static final int mod=1000000007;
	static final double eps=1e-8;
	static final long inf=100000000000000000L;
	static final boolean debug=true;
	static Reader in=new Reader();
	static StringBuilder ans=new StringBuilder();
	static long powm(long a,long b,long m){
		long an=1;
		long c=a;
		while(b>0){
			if(b%2==1)
				an=(an*c)%m;
			c=(c*c)%m;
			b>>=1;
		}
		return an;
	}
	static Random rn=new Random();
	static void sop(Object a){System.out.println(a);}
	static int ni(){return in.nextInt();}
	static int[] nia(int n){int a[]=new int[n];for(int i=0;i<n;i++)a[i]=ni();return a;}
	static long nl(){return in.nextLong();}
	static long[] nla(int n){long a[]=new long[n];for(int i=0; i<n; i++)a[i]=nl();return a;}
	static String ns(){return in.next();}
	static String[] nsa(int n){String a[]=new String[n];for(int i=0; i<n; i++)a[i]=ns();return a;}
	static double nd(){return in.nextDouble();}
	static double[] nda(int n){double a[]=new double[n];for(int i=0;i<n;i++)a[i]=nd();return a;}
	static class Reader{
		public BufferedReader reader;
		public StringTokenizer tokenizer;
		public Reader(){
			reader=new BufferedReader(new InputStreamReader(System.in),32768);
			tokenizer=null;
		}
		public String next(){
			while(tokenizer==null || !tokenizer.hasMoreTokens()){
				try{
					tokenizer=new StringTokenizer(reader.readLine());
				}
				catch(IOException e){
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}
		public int nextInt(){
			return Integer.parseInt(next());
		}
		public long nextLong(){
			return Long.parseLong(next());
		}
		public double nextDouble(){
			return Double.parseDouble(next());
		}
	}
}