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.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class A {
   
   static void print(ArrayList curr){
    
    for(int i=0;i<curr.size();i++)
        System.out.print(curr.get(i)+" ");
    System.out.println();
   }
    public static ArrayList mincost(int n,int k){
        if(k==0){
            ArrayList<Integer> ans = new ArrayList<Integer>();
            for(int i=1;i<=n;i++)
                ans.add(i);
            return ans;    
        }
        int  min = 1000000;
        ArrayList<Integer> ans = new ArrayList<Integer>();
        for(int s=0;s<=k;s++){
            // System.out.println("s= "+s);
            int count =0;
            ArrayList<Integer> curr = new ArrayList<Integer>(); 
            for(int i=1+s;i<=n;i=i+2*k+1){        //start planting at i+1
                count++;
                curr.add(i);
                if(i+2*k+1>n && i+k<n)
                    break;
                if(i+2*k+1>n && i+k>=n){
                    // System.out.println("cost= "+count);
                    // print(curr);
                    if(min>count){
                        min = count;
                        ans = (ArrayList<Integer>)curr.clone();
                        // for(int j=0;j<curr.size();j++){
                        //     ans.set(j,curr.get(j));
                        // }
                    }
                }
            }
        }
        return ans;
    }
    public static void main(String args[] ) throws Exception {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();
        ArrayList<Integer> ans = mincost(n,k); 
        System.out.println(ans.size());
        for(int i=0;i<ans.size();i++){
            System.out.print(ans.get(i)+" ");
        }
    }
}