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.Scanner;

/*https://codeforces.com/problemset/problem/1040/B
*/

public class ShashlikCookingProblem {

	/*
	 * n the number of skewers k number of skewers from each side that are turned in
	 * one step.
	 */

	public void solve(int n, int k) {

		int skewersCanBeTurnedOver = 2 * k + 1;
		int nextSkewerTobeTurned = 0;
		int minimumTurns = (n - 1) / skewersCanBeTurnedOver + 1;
		System.out.println(minimumTurns);
		if (k < n) {
			if (n % skewersCanBeTurnedOver > k || n % skewersCanBeTurnedOver == 0) {
				for (int i = 0; i < minimumTurns; i++) {
					nextSkewerTobeTurned = (k + 1) + i * (2 * k + 1);
					System.out.print(nextSkewerTobeTurned + " ");
				}
			} else {
				for (int i = 0; i < minimumTurns; i++) {
					nextSkewerTobeTurned = n % skewersCanBeTurnedOver + i * (2 * k + 1);
					System.out.print(nextSkewerTobeTurned + " ");
				}
			}
		} else {
			System.out.print(n + " ");
		}
	}

	public void solve1(int n, int k) {

		int skewersCanBeTurnedOver = 2 * k + 1;
		int nextSkewerTobeTurned = 0;
		int minimumTurns = (n - 1) / skewersCanBeTurnedOver + 1;
		System.out.println(minimumTurns);
		if (k < n) {

			if (n % skewersCanBeTurnedOver == 0) {
				for (int i = 0; i < minimumTurns; i++) {
					nextSkewerTobeTurned = (k + 1) + i * (2 * k + 1);
					System.out.print(nextSkewerTobeTurned + " ");
				}
			} else {
				for (int i = 0; i < minimumTurns; i++) {
					nextSkewerTobeTurned = n % skewersCanBeTurnedOver + i * (2 * k + 1);
					System.out.print(nextSkewerTobeTurned + " ");
				}
			}

		} else {
			System.out.print(n + " ");
		}
	}

	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);
		ShashlikCookingProblem shashlikCookingProblem = new ShashlikCookingProblem();
		int N = sc.nextInt();
		int K = sc.nextInt();

		shashlikCookingProblem.solve(N, K);

	}
}