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;

/**
 * @author hemenghai
 * @date 2019-01-11
 */
public class Main {

    private static int m, n, p, q;
    private static int cur, pos, sum, len;
    private static int max, min, ans, res;
    private static int ai, bi, ci, di;
    private static int[] aix, bix;
    private static long al, bl, cl, dl;
    private static long[] alx, blx;
    private static double ad, bd, cd, dd;
    private static double[] adx, bdx;
    private static boolean ab, bb, cb, db;
    private static boolean[] abx, bbx;
    private static char ac, bc, cc, dc;
    private static char[] acx, bcx;
    private static String as, bs, cs, ds;
    private static Scanner in = new Scanner(System.in);

    public static void main(String[] args) {
        m = in.nextInt();
        n = in.nextInt();

        min = n + 1;
        max = n * 2 + 1;
        ans = m / max + (m % max > 0 ? 1 : 0);
        ai = m % max;
        if(ai >= min){
            bi = ai - min + 1;
        } else if (ai == 0){
            bi = min;
        } else {
            bi = 1;
        }
        System.out.println(ans);
        for(;bi <= m;bi+=max){
            System.out.printf("%d ", bi);
        }
    }
}