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 r507_b extends Template{
    public static void main(String[] args) {
        int n = nI(), k = nI();
        ArrayList<Integer> al = new ArrayList<Integer>();
        int first = 0;
        if( (n%(2*k+1)) < (k+1) && (n%(2*k+1)!=0)){
            //first = (k+1) - (n%(2*k+1));
            first = (n%(2*k+1));
            al.add( first );
            first += k;
            n -= first;
        }
        while(n >= (2*k+1)){
            first += (k+1);
            al.add( first );
            first += k;
            n -= (2*k+1);
        }
        if(n>0){
            first += (k+1);
            al.add( first );
        }
        pn(al.size());
        for(int i=0; i<al.size(); i++){
            p(al.get(i));
        }
    }
}
 
class Template{
    static Scanner s = new Scanner(System.in);
    final static int nI(){return s.nextInt();}
    final static long nL(){return s.nextLong();}
    final static String nS(){return s.nextLine();}
    /* Renaming of some generic utility classes */
    final static class mylist extends ArrayList<pair>{}
    /* Constants Section */
    final static int INT_MIN = Integer.MIN_VALUE ;
    final static int INT_MAX = Integer.MAX_VALUE ;
    final static long LONG_MIN = Long.MIN_VALUE ;
    final static long LONG_MAX = Long.MAX_VALUE ;
    static long MOD = 1000000007 ;
    static void p(){System.out.println();}
    static void pn(long a){System.out.println(a);}
    static void pn(mylist a){System.out.println(a);}
    static void p(long a){System.out.printf("%d ", a);}
    static void p(long a, long b){System.out.printf("%d %d\n", a, b);}
    static void p(long a, long b, long c){System.out.printf("%d %d %d\n", a, b, c);}
    /* Implementation of operations modulo MOD */
    static long inv(long a){ return powM(a,MOD-2) ; }
    static long m(long a,long b){ return (a*b)%MOD ; }
    static long d(long a,long b){ return (a*inv(b))%MOD ; }
    static long powM(long a,long b){
        if(b<0) return powM(inv(a),-b) ;
        long ans=1 ;
        for( ; b!=0 ; a=(a*a)%MOD , b/=2) if((b&1)==1) ans = (ans*a)%MOD ;
        return ans ;
    }
}
class pair implements Comparable<pair>{
    long x ; long y ;
    pair(long _x,long _y){ x=_x ; y=_y ;}
    public int compareTo(pair p){
        return (this.x<p.x ? -1 : (this.x>p.x ? 1 : (this.y<p.y ? -1 : (this.y>p.y ? 1 : 0)))) ;
    }
    public String toString(){ return "(" + x + " " + y+ ")";}
}