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.io.*;
import java.util.*;
public class GFG {
	public static void main (String[] args) {
	Scanner sc=new Scanner(System.in);
	int n=sc.nextInt();
	int k=sc.nextInt();
	int flag=0;
	if(k!=0)
   { int r=n%(2*k+1);
      //System.out.println(r);
    int count1=(n/(2*k+1))+1;
    
	if(r!=0){
	     System.out.println(count1);
	      flag=1;
	      r=((r-1)/2)+1;
	          while(r<=n){
	             System.out.print(r+" ");  
	             r+=2*k+1;
	          }
	}
	else { r=k+1;
	 System.out.println(count1-1);
	     while(r<=n){
	             System.out.print(r+" ");  
	             r+=2*k+1;
	          }
	   // System.out.println(n);
	    
	}
   }
   else{
       System.out.println(n);
       for(int i=1;i<=n;i++)
       System.out.print(i+" ");
   }
	           
    //System.out.println(flag);	           
	}	          
}