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.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import javafx.util.Pair;

public  class CodeForce {
static	BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
static StringBuilder sb = new StringBuilder();

	static boolean flag = false;
	 
	static int next(String [] sr,int ind)
	{
		return Integer.parseInt(sr[ind]);
	}

	public static void main(String[] args) throws IOException {
		
String[] sr=br.readLine().split(" ");
int n=next(sr,0);
int k=next(sr,1);

int y=n%(2*k+1);
if(y!=0&&y<=k+1)
{
System.out.println(n/(2*k+1)+1);

int i=1;
while(i<=n) {
	System.out.print(i+" ");
	i=i+2*k+1;
	
}
return;
	}
else
{
	if(y==0)
		System.out.println(n/(2*k+1));
	else
		System.out.println(n/(2*k+1)+1);
		
		
		
		int i=k+1;
		while(i<=n)
			{System.out.print(i+" ");i=i+2*k+1;}
		
		return;
		}
	
}




		
		
		
		
		
		
		
		
		

		
		
	}