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 Main {
 public static void main(String args[])
 {
	 Scanner input=new Scanner(System.in);
	 int n,k;int num=0;
	 n=input.nextInt();k=input.nextInt();
	 if(n<=2*k+1)
	 {
		 System.out.println(1);
		 System.out.println((1+n)/2);
	 }
	 else
	 {
		 int index=0;int fnum=0;
	 for(int i=2*k+1;i>=k+1;i--)
	 {
		 int sub=n-i;
		 int mod=sub%(2*k+1);
		 if(mod==0)
		 {
			 num+=1+sub/(2*k+1);
			 fnum=i;break;
		 }
		 else if(mod>=k+1)
		 {
			 num+=1+sub/(2*k+1);
			 if(sub%(2*k+1)!=0)num+=1;
			 fnum=i;
			 break;
		 }
		
	 }
	 System.out.println(num);
	 System.out.print(fnum-k+" ");
	 index=fnum-k;
	 for(int i=1;i<=num-1;i++)
	 {
		 System.out.print(index+2*k+1+" ");
		 index+=2*k+1;
	 }
 }
}
}