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.*;import java.math.*;

public class First
{

    public static ArrayList<Integer> getAns(int m, int in, int shift,int y,int n)
    {
         ArrayList<Integer> list = new ArrayList<Integer>();
         in-=shift;
         for(int i=0;i<m;i++)
         {
              list.add(in);
              in+=y;
         }
         return list;
    }
    public static void process()throws IOException
    {
         int n = ni();
         int k = ni();
         int y = (2*k)+1;
         int count=0;
         ArrayList<Integer> list = new ArrayList<Integer>();
         if(k>=((n+1)/2))
         {
              count=1;
              list.add((n+1)/2);
         }
         else
         {
               int m = n/y;
               int in = (y+1)/2;
               int left = n%y;;
               if(left==0)
               {
                    list = getAns(m,in,0,y,n);
               }
               else
               {
                    if(left<=k+1)
                    {
                         list = getAns(m,in,k+1-left,y,n);
                         list.add(n);
                    }
                    else
                    {
                          list = getAns(m,in,0,y,n);
                          int h = list.get(list.size()-1);
                          list.add(h+(2*k)+1);
                    }
               }
               count=list.size();
         }
         pn(count);
         StringBuilder sb = new StringBuilder("");
         for(Integer i : list)
         {
              sb.append(i).append(" ");
         }
         pn(sb.toString());
    }


   	static AnotherReader sc;
    static PrintWriter out;
    public static void main(String[]args)throws IOException
    {
        out = new PrintWriter(System.out);
        sc=new AnotherReader();
        boolean oj = true;

    // 	oj = System.getProperty("ONLINE_JUDGE") != null;
    // 	if(!oj) sc=new AnotherReader(100);

    //     long s = System.currentTimeMillis();
        int t=1;
        while(t-->0)
            process();
        out.flush();
        // if(!oj)
        //     System.out.println(System.currentTimeMillis()-s+"ms");
        System.out.close();  
    }

    static void pn(Object o){out.println(o);}
    static void p(Object o){out.print(o);}
    static void pni(Object o){out.println(o);System.out.flush();}
    static int ni()throws IOException{return sc.nextInt();}
    static long nl()throws IOException{return sc.nextLong();}
    static double nd()throws IOException{return sc.nextDouble();}
    static String nln()throws IOException{return sc.nextLine();}
    static long gcd(long a, long b)throws IOException{return (b==0)?a:gcd(b,a%b);}
    static int gcd(int a, int b)throws IOException{return (b==0)?a:gcd(b,a%b);}
    static int bit(long n)throws IOException{return (n==0)?0:(1+bit(n&(n-1)));}
    static boolean multipleTC=false;



/////////////////////////////////////////////////////////////////////////////////////////////////////////

    static class AnotherReader{BufferedReader br; StringTokenizer st;
    AnotherReader()throws FileNotFoundException{
    br=new BufferedReader(new InputStreamReader(System.in));}
    AnotherReader(int a)throws FileNotFoundException{
    br = new BufferedReader(new FileReader("input.txt"));}
    String next()throws IOException{
    while (st == null || !st.hasMoreElements()) {try{
    st = new StringTokenizer(br.readLine());}
    catch (IOException  e){ e.printStackTrace(); }}
    return st.nextToken(); } int nextInt() throws IOException{
    return Integer.parseInt(next());}
    long nextLong() throws IOException
    {return Long.parseLong(next());}
    double nextDouble()throws IOException { return Double.parseDouble(next()); }
    String nextLine() throws IOException{ String str = ""; try{
    str = br.readLine();} catch (IOException e){
    e.printStackTrace();} return str;}}
    
/////////////////////////////////////////////////////////////////////////////////////////////////////////////
}