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.ArrayList;
import java.util.Scanner;

public class ShashlikCooking1 {
	
	public static void main(String[] args) {
		
		Scanner sc = new Scanner (System.in) ;
		
		String s = sc.nextLine() ;
		
		String [] priceandlength = s.split("\\s+") ;
		
		int n = Integer.parseInt(priceandlength[0]) ;
		int k = Integer.parseInt(priceandlength[1]) ;
		
		
		int formula = 1 + (k*2) ;

		boolean found = false ;
		
		ArrayList<Integer> a = null ;
		
		for(int i = 1 ; i <= 1+k ; i ++)
		{
			a = new ArrayList() ;
			
			a.add(i) ;
			
			for(int j = i + formula ; j <= n ; j += formula)
			{
				a.add(j) ;
				if(n - j  <= k)
				{
					found = true ;
					break ;
				}
			}
			
			if(found)
				break ;
			
		}
		
		System.out.println(a.size()) ;
		if(found)
		{
			for(int i = 0 ; i < a.size() ; i ++)
			{
				System.out.print(a.get(i) + " ") ;
			}
		}
		else
		{
			if(n%2 != 0)
				System.out.println((n/2) + 1) ;
			else
				System.out.println(n/2) ;
		}
		
	}

}