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.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class Problem507B {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		out=new PrintWriter(new BufferedOutputStream(System.out));
		FastReader s=new FastReader();
		int n=s.nextInt();
		int k=s.nextInt();
		if(k==0) {
			out.println(n);
			for(int i=1;i<=n;i++) {
				out.print(i+" ");
			}
			out.close();
			System.exit(0);
		}
		if(n<=2*k+1) {
			out.println(1);
			out.println(n/2+1);
			out.close();
			System.exit(0);
		}
		if(n%(2*k+1)==0) {
			out.println(n/(2*k+1));
			int count=1;
			for(int i=k+1;i<=n&&count<=n/(2*k+1);i+=2*k+1) {
				out.print(i+" ");
				count++;
			}
			out.close();
			System.exit(0);
		}
		int current=n/(2*k+1)-1;
		int remaining=n%(2*k+1)+2*k+1;
		int first=1;
		int second=n;
		for(int i=2*k+1;i>=k+1;i--) {
			int remain=remaining-i;
			if(remain>=k+1 && remain<=2*k+1) {
				first=i-k;
				second=n+1-remain+k;
				break;
			}
		}
		out.println(current+2);
		;
		int i=first;
		for(i=first;i+2*k+1<=n;i+=2*k+1) {
			out.print(i+" ");
		}
		if(i+2*k+1>n)out.print(second);
		out.close();
	}

	public static PrintWriter out;
	
	public static class FastReader {
		BufferedReader br;
		StringTokenizer st;
		//it reads the data about the specified point and divide the data about it ,it is quite fast
		//than using direct 

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (Exception e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());//converts string to integer
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (Exception e) {
				e.printStackTrace();
			}
			return str;
		}
	} 
	
}