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.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.List;
import java.util.StringTokenizer;

public class Main {

	public static void main(String[] args) throws IOException {
		int n = IO.nextInt();
		int k = IO.nextInt();
		
		List<Integer> s = new ArrayList<>(n);
		for(int i = 0; i < n;) {
			i += k;
			i++;
			s.add(i);
			i += k;
		}
		
		IO.write(s.size() + "\n");
		int last = s.get(s.size() - 1);
		int dif = 0;
		if(last > n)
			dif = last - n;
		for(int i : s)
			IO.write(i - dif + " ");
		IO.write("\n");
		IO.dispose();
	}

	static class IO {
		private static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		private static BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
		private static StringTokenizer token = new StringTokenizer("");
			
		static String readLine() throws IOException {
			return br.readLine();
		}
			
		static String next() throws IOException {
			while(!token.hasMoreTokens())
				token = new StringTokenizer(br.readLine(), " ");
			return token.nextToken();
		}
		
		static int nextInt() throws IOException {
			return Integer.parseInt(next());
		}
		
		static long nextLong() throws IOException {
			return Long.parseLong(next());
		}
		
		static BigInteger nextBigInteger() throws IOException {
			return new BigInteger(next());
		}
		
		static double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}
		
		static void write(String str) throws IOException {
			bw.write(str);
		}
		
		static void dispose() throws IOException {
			br.close();
			bw.close();
		}
	}
}