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.*;
public class Main {


    public static void main (String[] args) {
    	Scanner s=new Scanner(System.in);
      int n=s.nextInt(),k=s.nextInt();
      int c=Integer.MAX_VALUE;
      ArrayList<Integer> a=new ArrayList<>();
      boolean d=true;
      if(n<=2*k+1) {
    	  System.out.println(1);
    	  System.out.println(n/2+1);
    	  return;
      }
     
   for(int i=1;i<=k+1;i++) {
    	 ArrayList<Integer> x=new ArrayList<>();
         int j=i;
    	 int l=0;
    	 boolean d1=true;
    	 while(j<=n) {
    		 l++;
    		 x.add(j);
    		 if(j+2*k+1>n) {
    			 if(j+k+1<=n) {
    			
    				 d1=false;}
    			
    				 j=j+2*k+1;
    		 }
    		 else
    			 j=j+2*k+1;
    	 }
    	 if(l<c&&d1) {
    		 c=l;
    		 while(a.size()!=0) {
    			 a.remove(a.size()-1);
    		 }
    		 d=false;
    		 for(int m=0;m<x.size();m++)
    			 a.add(x.get(m));
    	 }
   }
     
     System.out.println(c);
     for(int i=0;i<a.size();i++)
    	 System.out.print(a.get(i)+" ");
    }
}