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.* ;


public class JavaApplication8 {

    
    public static void main(String[] args) {
          Scanner in = new Scanner(System.in) ;
          int N = in.nextInt() ;
          int k = in.nextInt() ;
          
          if( N % (2*k+1) == 0){
              System.out.println( N / (2*k +1)) ;
              
              System.out.print((1+k)+" ") ;
              
              int cur = k+1; 
              for(int i=1;i<=N / (2*k +1)-1;i++){
                  System.out.print((cur+2*k+1)+" ") ;
                  cur += 2*k+1;
              }
              
          }else{
              double Nf = (double) N ;
              double Kf = (double) k ;
              
              int n = (int) Math.ceil( Nf / (2*Kf+1) ) - 2 ;
          
          if(n < 0){
              System.out.println(1+"\n"+(N/2+1));
          }else{
              System.out.println(n+2);
              
              int s =  N - (2*k+1)*(n+1)  - 1 ;
              int x = s / 2 ;
              int y = s - x ;
              
              System.out.print((1+x)+" ") ;
              
              int cur = x+1 ;
              for(int i=1;i<=n;i++){
                  System.out.print((cur+2*k+1)+" ") ;
                  cur = cur + 2*k+1 ;
              }
              
              System.out.print(N -y) ;
          }
        }
          
                  
    }
}