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:
//package Main;

import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.Iterator;
import java.util.Scanner;

public class Main {
	
	static int maxn = 1010; 
	static int[] p = new int[maxn];
	public static void main(String[] args) {
		@SuppressWarnings("resource")
		Scanner input = new Scanner(System.in);
		int n = input.nextInt();
		int k = input.nextInt();
		if(2*k+1>=n){
			System.out.println("1");
			System.out.println((n+1)/2);
			return;
		}
		int cnt=0,pos=k+1;
		while(pos-k<=n){
			p[++cnt]=pos;
			//System.out.println("pos= "+pos);
			pos+=(2*k+1);
		}
		int d=0;
		if(p[cnt]>n) d=p[cnt]-n;
		for(int i=1;i<=cnt;i++) p[i]-=d;
		System.out.println(cnt);
		for(int i=1;i<=cnt;i++){
			if(i==1) System.out.print(p[1]);
			else System.out.print(" "+p[i]);
		}
		System.out.println();
	}
}