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.*;

public class Main {


    public static void main(String[] args) {

        Scanner s = new Scanner(System.in);
        int n = s.nextInt(), k = s.nextInt();
        ArrayList<Integer> res = new ArrayList<>();

        if (n <= (2 * k + 1)) {
            System.out.println(1);
            System.out.println(n / 2 + 1);
            return;
        }
        if (k == 0) {
            System.out.println(n);
            for (int i = 1; i <= n; i++) {
                System.out.print(i + " ");
            }
            return;
        }
        int x, y;

        int totNumOfActions = 0;
        for (int i = k + 1; i <= 2 * k + 1; i++) {
            for (int j = 2 * k + 1; j >= k + 1; j--) {
                if ((n - i - j) >= 0 && (n - i - j) % (k * 2 + 1) == 0){
                    x = i;
                    y = j;
                    res.add(x - k);
                    totNumOfActions++;
                    for (int l = x; l < n - y; l += k * 2 + 1) {
                        res.add(l + k + 1);
                        totNumOfActions++;
                    }
                    res.add(n - y + k + 1);
                    totNumOfActions++;

                    System.out.println(totNumOfActions);
                    for (int l = 0; l < res.size(); l++) {
                        System.out.print(res.get(l) + " ");
                    }
                    return;
                }

            }

        }

    }
}