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.Scanner;
import java.util.Arrays;
import java.util.*;
import java.io.*;
public class Main507B
{
  static boolean[] a;
  static PrintWriter out=new PrintWriter(System.out);
  public static void main(String[] args) throws IOException
  {
    Reader sc=new Reader();
    int n=sc.nextInt();
    int k=sc.nextInt();
    boolean flag=false;
    a=new boolean[n+1];
    
    int x=(int)Math.ceil((double)n/(2*k+1));
    int ans[]=new int[x];
    int p=0;
    //System.out.println(x);
    for(int i=1;i<=n;i++)
    {
      Arrays.fill(a,false);
      int t=x;
      int j=i;
      while(t-->0)
      {
        process(j,k);
        if(check())
        {
          for(int l=0;l<x;l++)
            ans[l]=i+l*2*k+l;
          flag=true;
          break;
        }
        j=j+2*k+1;
      }
      if(flag)
        break;
    }
    if(flag)
    {
      out.println(x);
      for(int i=0;i<x;i++)
        out.print(ans[i]+" ");
    }
    out.flush();
  }
  public static void process(int i, int k)
  {
    int st=i-k;
    if(i-k<=0)
      st=1;
    int lo=i+k;
    if(i+k>a.length-1)
      lo=a.length-1;
    for(int l=st;l<=lo;l++)
      a[l]=!a[l];
  }
  public static boolean check()
  {
    for(int i=1;i<=a.length-1;i++)
      if(!a[i]) return false;
    return true;
  }
  static class Reader {
  final private int BUFFER_SIZE = 1 << 16;
  private DataInputStream din;
  private byte [] buffer;
  private int bufferPointer, bytesRead;
  
  public Reader () {
   din = new DataInputStream (System.in);
   buffer = new byte[BUFFER_SIZE];
   bufferPointer = bytesRead = 0;
  }
  
  public Reader (String file_name) throws IOException {
   din = new DataInputStream (new FileInputStream (file_name));
   buffer = new byte[BUFFER_SIZE];
   bufferPointer = bytesRead = 0;
  }
  
  public String readLine () throws IOException {
   byte [] buf = new byte[1024];
   int cnt = 0, c;
   while ((c = read ()) != -1) {
    if (c == '\n')
     break;
    buf[cnt++] = (byte) c;
   }
   return new String (buf, 0, cnt);
  }
  
  public int nextInt () throws IOException {
   int ret = 0;
   byte c = read ();
   while (c <= ' ')
    c = read ();
   boolean neg = (c == '-');
   if (neg)
    c = read ();
   do {
    ret = ret * 10 + c - '0';
   } while ((c = read ()) >= '0' && c <= '9');
   if (neg)
    return -ret;
   return ret;
  }
  
  public long nextLong () throws IOException {
   long ret = 0;
   byte c = read ();
   while (c <= ' ')
    c = read ();
   boolean neg = (c == '-');
   if (neg)
    c = read ();
   do {
    ret = ret * 10 + c - '0';
   } while ((c = read ()) >= '0' && c <= '9');
   if (neg)
    return -ret;
   return ret;
  }
  
  public double nextDouble () throws IOException {
   double ret = 0, div = 1;
   byte c = read ();
   while (c <= ' ')
    c = read ();
   boolean neg = (c == '-');
   if (neg)
    c = read ();
   do {
    ret = ret * 10 + c - '0';
   } while ((c = read ()) >= '0' && c <= '9');
   if (c == '.')
    while ((c = read ()) >= '0' && c <= '9')
     ret += (c - '0') / (div *= 10);
   if (neg)
    return -ret;
   return ret;
  }
  
  private void fillBuffer () throws IOException {
   bytesRead = din.read (buffer, bufferPointer = 0, BUFFER_SIZE);
   if (bytesRead == -1)
    buffer[0] = -1;
  }
  
  private byte read () throws IOException {
   if (bufferPointer == bytesRead)
    fillBuffer ();
   return buffer[bufferPointer++];
  }
  
  public void close () throws IOException {
   if (din == null)
    return;
   din.close ();
  }
 }
}