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;

public class CF507B {
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int n = s.nextInt();
        int k = s.nextInt();
        if(k == 0){
            System.out.println(n);
            for (int i = 1; i <= n; i++) {
                System.out.println(i);
            }
        }else {
            if (n <= k) {
                System.out.println(1);
                System.out.println(1);
            } else {
                if (n % (2 * k + 1) == 0) {
                    System.out.println(n / (2 * k + 1));
                    int j = 2 * k + 1;
                    k = k + 1;
                    while (k < n) {
                        System.out.print(k + " ");
                        k += j;
                    }
                } else {
                    System.out.println(n / (2 * k + 1) + 1);
                    int div = n % (2 * k + 1);
                    if (n % (2 * k + 1) <= k) {
                        System.out.print(1 + " ");
                        int j = 2 * k + 1;
                        int m = k + 1 + k + 1;
                        while (m < n - k - div) {
                            System.out.print(m + " ");
                            m += j;
                        }
                        System.out.print(n - div + 1);
                    } else {
//                    System.out.println(n / (2 * k + 1));
                        int j = 2 * k + 1;
                        k = k + 1;
                        while (k <= n) {
                            System.out.print(k + " ");
                            k += j;
                        }

                    }


                }
                System.out.println();
            }
        }
    }
}