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.File;
import java.io.FileNotFoundException;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintStream;
import java.util.*;
import java.util.StringTokenizer;

public class Tester {

	static class FastReader
    {
        BufferedReader br;
        StringTokenizer st;
 
        public FastReader()
        {
            br = new BufferedReader(new
                     InputStreamReader(System.in));
        }
 
        String next()
        {
            while (st == null || !st.hasMoreElements())
            {
                try
                {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException  e)
                {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
 
        int nextInt()
        {
            return Integer.parseInt(next());
        }
 
        long nextLong()
        {
            return Long.parseLong(next());
        }
 
        double nextDouble()
        {
            return Double.parseDouble(next());
        }
 
        String nextLine()
        {
            String str = "";
            try
            {
                str = br.readLine();
            }
            catch (IOException e)
            {
                e.printStackTrace();
            }
            return str;
        }
    }
	
	public static void main(String[] args) throws FileNotFoundException {
		// TODO Auto-generated method stub
		
		solveQ2();
		
	}

	private static void solveQ2() {
		// TODO Auto-generated method stub
		
		FastReader sc = new FastReader();
		
		int n = sc.nextInt();
		int k = sc.nextInt();
		
		HashMap<Integer, Integer> hm = new HashMap<>();
		hm.put(0,0);
		int min = Integer.MAX_VALUE;
		
		for(int i=1; i<=n; i++) {
		
			int prev = Math.max(i-k-1, 0);
			int next = Math.min(i+k, n);
			
			if(hm.containsKey(prev)) {
				hm.put(next, i);
			}
			
			if(hm.containsKey(n))
				break;
						
		}
		
		hm.remove(0);
		ArrayList<Integer> a = new ArrayList<Integer>();
		
		int temp = n;
		while(hm.containsKey(temp)) {
			
			a.add(hm.get(temp));
			temp = hm.get(temp)-k-1;
				
		}
		
		System.out.println(a.size());
		for(int i=a.size()-1; i>=0; i--)
			System.out.print(a.get(i) + " ");
		
	}

	private static void solveQ1() {
		// TODO Auto-generated method stub
		
		FastReader sc = new FastReader();
		int n = sc.nextInt();
		int x = sc.nextInt();
		int y = sc.nextInt();
		
		int a[] = new int[n];
		for(int i=0; i<n; i++)
			a[i] = sc.nextInt();
		
		int l = 0;
		int r = n-1;
		long ans = 0;
		while(l<r) {
			
			if(a[l]==2 && a[r] == 2) {
				ans += 2*Math.min(x, y);
			}
			else if(a[l] == 2) {
				if(a[r] == 0)
					ans += x;
				else
					ans+=y;
			}
			else if(a[r] == 2) {
				if(a[l] == 0)
					ans+=x;
				else
					ans+=y;
			}
			else {
				if(a[l]!=a[r]) {
					System.out.println(-1);
					return;
				}
			}
			
			l++;
			r--;
			
		}
	
		if(l == r && a[l] == 2)
			ans += Math.min(x, y);
		
		System.out.println(ans);
		
	}


}