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:
//package contese_476;
import java.util.*;
public class q1 
{
	int m=(int)1e9+7;
public class Node
{
	int a;
	int b;
	public void Node(int a,int b)
	{
		this.a=a;
		this.b=b;
	}
}
public int mul(int a ,int b)
{
a=a%m;
b=b%m;
return((a*b)%m);
}
public int pow(int a,int b)
{
	int x=1;
	while(b>0)
	{
		if(b%2!=0)
			x=mul(x,a);
		a=mul(a,a);
		b=b/2;
	}
	return x;
}
public static long gcd(long a,long b)
{
	if(b==0)
		return a;
	else
		return gcd(b,a%b);
}
public static void main(String[] args)
{
	Scanner sc=new Scanner(System.in);
	int n=sc.nextInt();
	int k=sc.nextInt();

		
	if(n<=(k+1))
	{
		System.out.println(1);
		System.out.print(n);
		System.exit(0);
	}
	if(n<=(2*k+1))
	{
		System.out.println(1);
		System.out.print(k+1);
		System.exit(0);
	}
	ArrayList<Integer> arr=new ArrayList();
	int num=n/(2*k+1);
	int r=n%(2*k+1);
	int n1=(2*k+1)+r;
	//System.out.print(n1+" ");
	if(r==0)
	{
		System.out.println(num);
		int i=n-k;
		int count=0;
		while(count<num)
		{
			arr.add(i);
			i-=(2*k+1);
			count++;
		}
		for(int j=arr.size()-1;j>=0;j--)
			System.out.print(arr.get(j)+" ");
	}
	if(r!=0)
	{
	System.out.println(num+1);
	int x=2*k+1;
	int count=0;
	int sum=0;
	while(count<num-1)
	{
		sum+=(2*k+1);
		count++;
	}
	n1=(n-sum);
	int a,b;
      if(n1%2==0)
      {
    	  a=n1/2;
    	  b=n1/2;
      }
      else
      {
    	  a=n1/2;
    	  b=(n1/2)+1;
      }
      int i=(a+k+1);
      a=(2*k+1)-a;
      b=(2*k+1)-b;
      System.out.print(((k+1)-a)+" ");
      count=0;
      while(count<num-1)
      {
    	  System.out.print(i+" ");
    	  i+=(2*k+1);
    	  count++;
      }
      System.out.print(((n-k)+b)+" ");
	
	}
	
	
     
}

}