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.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.StringTokenizer;

public class B {
  public static void main(String[] args)throws Exception {
	BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	StringTokenizer tk = new StringTokenizer(br.readLine());
	int n = Integer.parseInt(tk.nextToken());
	int k = Integer.parseInt(tk.nextToken());
	
	PrintWriter out = new PrintWriter(System.out);
	
	if(2*k+1 >=n)
	{
		out.println(1);
		out.println((n+1)/2);
		
	}
	else
	{
		
		ArrayList<Integer> ans = new ArrayList<>();
        int rem =2*k+1-n%(2*k+1);
        
        if (rem == 2*k +1)
            rem = 0;
        
        if(rem>k)
        	rem=k;
        
        for(int i =k+1-rem;i<=n;i+=2*k+1)
        	ans.add(i);

        out.println(ans.size());
        for (int x : ans)
            out.print(x + " ");


	}
	
	out.flush();
	out.close();
}
}