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.stream.Stream;

/**
 * @author madi.sagimbekov
 */
public class C1040B {

    private static BufferedReader in;
    private static BufferedWriter out;

    public static void main(String[] args) throws IOException {
        open();

        int[] x = readInts();
        int n = x[0];
        int k = x[1];

        int rem = n % (2 * k + 1);

        if (rem > k) {
            int cnt = n / (2 * k + 1) + 1;
            out.write(cnt+"\n");
            int start = k + 1;
            for (int i = 0; i < cnt; i++) {
                out.write(start + " ");
                start += 2 * k + 1;
            }
        } else if (rem > 0) {
            int p = (2 * k + 1 + rem);
            int a = (p + 1)/ 2;
            int b = p / 2;

            int start = a - k;
            int cnt = n / (2 * k + 1) + 1;
            out.write(cnt+"\n");
            for (int i = 0; i < cnt; i++) {
                out.write(start + " ");
                start += 2 * k + 1;
            }
            out.write("\n");
        } else {
            int cnt = n / (2 * k + 1);
            int start = k + 1;
            out.write(cnt+"\n");
            for (int i = 0; i < cnt; i++) {
                out.write(start + " ");
                start += 2 * k + 1;
            }
            out.write("\n");
        }

        close();
    }

    private static int[] readInts() throws IOException {
        return Stream.of(in.readLine().split(" ")).mapToInt(Integer::parseInt).toArray();
    }

    private static int readInt() throws IOException {
        return Integer.parseInt(in.readLine());
    }

    private static long[] readLongs() throws IOException {
        return Stream.of(in.readLine().split(" ")).mapToLong(Long::parseLong).toArray();
    }

    private static long readLong() throws IOException {
        return Long.parseLong(in.readLine());
    }

    private static double[] readDoubles() throws IOException {
        return Stream.of(in.readLine().split(" ")).mapToDouble(Double::parseDouble).toArray();
    }

    private static double readDouble() throws IOException {
        return Double.parseDouble(in.readLine());
    }

    private static String readString() throws IOException {
        return in.readLine();
    }

    private static void open() {
        in = new BufferedReader(new InputStreamReader(System.in));
        out = new BufferedWriter(new OutputStreamWriter((System.out)));
    }

    private static void close() throws IOException {
        out.flush();
        out.close();
        in.close();
    }

}