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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.FilterInputStream;
import java.io.BufferedInputStream;
import java.util.ArrayList;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author nirav
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scan in = new Scan(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        BShashlikCooking solver = new BShashlikCooking();
        solver.solve(1, in, out);
        out.close();
    }

    static class BShashlikCooking {
        public void solve(int testNumber, Scan in, PrintWriter out) {
            int n = in.scanInt();
            int k = in.scanInt();
            if (n <= (2 * k + 1)) {
                out.println("1");
                out.println((n + 1) / 2);
                return;
            }
            int m = n % (2 * k + 1);
            ArrayList<Integer> array = new ArrayList<>();
            if (m != 0) {
                if (m > k) {
                    int ans = m - k;
                    while (ans <= n) {
                        array.add(ans);
                        ans += (2 * k + 1);
                    }
                } else {
                    int ans = 1;
                    while (ans <= n) {
                        array.add(ans);
                        ans += (2 * k + 1);
                    }
                }

                out.println(array.size());
                for (int i = 0; i < array.size(); i++) out.print(array.get(i) + " ");
                out.println();
            } else {
                int ans = k + 1;
                out.println(n / (2 * k + 1));
                while (ans <= n) {
                    out.print(ans + " ");
                    ans += (2 * k + 1);
                }
                return;
            }
        }

    }

    static class Scan {
        private byte[] buf = new byte[4 * 1024];
        private int INDEX;
        private BufferedInputStream in;
        private int TOTAL;

        public Scan(InputStream inputStream) {
            in = new BufferedInputStream(inputStream);
        }

        private int scan() {
            if (INDEX >= TOTAL) {
                INDEX = 0;
                try {
                    TOTAL = in.read(buf);
                } catch (Exception e) {
                    e.printStackTrace();
                }
                if (TOTAL <= 0) return -1;
            }
            return buf[INDEX++];
        }

        public int scanInt() {
            int I = 0;
            int n = scan();
            while (isWhiteSpace(n)) n = scan();
            int neg = 1;
            if (n == '-') {
                neg = -1;
                n = scan();
            }
            while (!isWhiteSpace(n)) {
                if (n >= '0' && n <= '9') {
                    I *= 10;
                    I += n - '0';
                    n = scan();
                }
            }
            return neg * I;
        }

        private boolean isWhiteSpace(int n) {
            if (n == ' ' || n == '\n' || n == '\r' || n == '\t' || n == -1) return true;
            else return false;
        }

    }
}