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.*;
import static java.lang.Math.*;
import java.io.*;

public class SolutionB {
       
      public static void main(String args[])throws IOException{
              Scanner sc = new Scanner(System.in);
              int n = sc.nextInt();
              int k = sc.nextInt();
              int max = Integer.MAX_VALUE;
              List<Integer> maxL = new ArrayList<Integer>();
              for(int i = 1; i <= k+1; i++){
                     List<Integer> curr = new ArrayList<Integer>();
                     int j = i;
                     boolean possible = false;
                     while(j <= n){
                            curr.add(j);
                            if((j + k) >= n)
                                   possible = true;
                            j += (2*k+1);
                     }
                     if(possible && curr.size() < max){
                            max = curr.size();
                            maxL = curr;
                     }
              }
              System.out.println(max);
              for(int i = 0; i < maxL.size(); i++)
                     System.out.print(maxL.get(i) + " ");
              System.out.println();
      }
}