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.io.*;
import java.util.*;

public class B {
    String fileName = "";

    public void solve() throws IOException {
        int n = nextInt();
        int k = nextInt();
        int cnt = 2 * k + 1;
        List<Integer> ans = new ArrayList<>();
        int l = 1;
        int r = n;
        int mod = n % cnt;
        if (mod != 0 && k + 1 > mod) {
            mod -= k + 1;
            mod += cnt;
            mod %= cnt;
            ans.add(1);
            l += k + 1;
        }
        if (l <= r) {
            if (mod != 0) {
                r = n - mod;
                ans.add(n - (mod - k - 1));
            }
            for (int i = l + k; i <= r; i += cnt) {
                ans.add(i);
            }
        }
        out.println(ans.size());
        for (int i = 0; i < ans.size(); i++) {
            out.print(ans.get(i));
            out.print(' ');
        }
    }

    public void run() throws IOException {
        br = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        solve();
        br.close();
        out.close();
    }

    BufferedReader br;
    StringTokenizer in;
    PrintWriter out;

    public String nextToken() throws IOException {
        while (in == null || !in.hasMoreTokens()) {
            in = new StringTokenizer(br.readLine());
        }
        return in.nextToken();
    }

    public int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    public double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }

    public long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    public static void main(String... args) throws IOException {
        new B().run();
    }
}