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.ArrayList;
import java.util.List;
import java.util.Scanner;

/**
 * @author Sergey Skrobotov
 */
@SuppressWarnings("all")
public class Solution {

    private static Scanner scanner = new Scanner(System.in);

    public static final int MOD = 1000000007;

    public static final boolean WITH_TEST_CASES = false;

    public static void main(String[] argv) throws Exception {
        solution();
    }

    public static void solution() {
        final Solution s = new Solution();
        int t = WITH_TEST_CASES ? scanner.nextInt() : 1;
        for (int i = 0; i < t; i++) {
            s.start();
        }
    }

    public void start() {
        final int n = scanner.nextInt(); // 1..1000
        final int k = scanner.nextInt(); // 0..1000
        final List<Integer> result = new ArrayList<>();
        int offset = 0;
        int nextUnset = 0;
        while (true) {
            final int i = nextUnset + k;
            if (i >= n) {
                offset = n - 1 - i;
            }
            result.add(i);
            nextUnset = i + k + 1;
            if (nextUnset >= n) {
                break;
            }
        }
        System.out.println(result.size());
        final int off = offset;
        result.forEach(q -> System.out.print((q + off + 1) + " "));
        System.out.println();
    }
}