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.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

/**
 * 
 */

/**
 * @author timat
 *
 */
public class Task {

	/**
	 * @param args
	 * @throws IOException
	 */
	public static void main(String[] args) throws IOException {
		FastIO.init();

		int n = FastIO.readInt();
		int k = FastIO.readInt();

		int result = Math.max(1, n / (2 * k + 1) + (n % (2 * k + 1) == 0 ? 0 : 1));

		System.out.println(result);
		for (int i = ((n % (2 * k + 1)) == 0 ? k + 1 : Math.min(k + 1, Math.max(1, (n % (2 * k + 1))))); i <= n; i += (2 * k + 1)) {
			System.out.print(i + " ");
		}
		System.out.println();

		FastIO.cleanup();
	}

	private static class FastIO {
		private static StringTokenizer tokenizer;
		private static BufferedReader in;

		public static void init() {
			in = new BufferedReader(new InputStreamReader(System.in));
		}

		public static void initf() throws FileNotFoundException {
			initf("in.txt");
		}

		public static void initf(String name) throws FileNotFoundException {
			File file = new File(name);
			in = new BufferedReader(file.exists() ? new FileReader(file) : new InputStreamReader(System.in));
		}

		public static void cleanup() throws IOException {
			in.close();
		}

		public static String readToken() throws IOException {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				tokenizer = new StringTokenizer(in.readLine());
			}
			return tokenizer.nextToken();
		}

		public static int readInt() throws NumberFormatException, IOException {
			return Integer.parseInt(readToken());
		}

	}

}