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.math.BigInteger;
import java.util.*;
import java.io.*;
import java.text.*;

public class Main {
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in), 32768);
    static PrintWriter out = new PrintWriter(System.out);
    static StringTokenizer t;

    static String sn() {
        while (t == null || !t.hasMoreTokens()) {
            try {
                t = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                throw new RuntimeException();
            }
        }
        return t.nextToken();
    }

    static int ni() {
        return Integer.parseInt(sn());
    }

    static long nlo() {
        return Long.parseLong(sn());
    }

    public static void main(String args[]) {
        int t = 1;
        while (t-- > 0) {
            solve();
        }
        out.close();
    }

    static long mod = (long) 998244353;

    static long pro(long a, long b) {
        return (a % mod * b % mod) % mod;
    }

    static long add(long a, long b) {
        return (a + b) % mod;
    }

    static long sub(long a, long b) {
        return ((a % mod - b % mod) + mod) % mod;
    }

    static int maxn = 1000005;

    static int pow(int n) {
        return (int) (Math.log10(n) / Math.log10(2));
    }

    static long modpow(long x, long y) {
        long res = 1l;
        x %= mod;
        while (y > 0) {
            if (y % 2 != 0)
                res = pro(res, x);
            x = pro(x, x);
            y /= 2;
        }
        return res;
    }

    static class pair {
        String a;
        int b;

        pair(String x, int y) {
            a = x;
            b = y;
        }
    }

    static long modInverse(long n) {
        BigInteger n1 = new BigInteger(Long.toString(n));
        String s = (n1.modInverse(new BigInteger(Long.toString(mod)))).toString();
        return Long.parseLong(s);
    }

    static boolean v[] = new boolean[1000005];

    static void seive() {
        v[1] = true;
        for (int i = 2; i < 1000005; i++) {
            if (!v[i]) {
                for (int j = 2 * i; j < 1000005; j += i)
                    v[j] = true;
            }
        }
    }

    static boolean isprime(long n) {
        if (n == 1)
            return false;
        for (int i = 2; i * i <= n; i++) {
            if (n % i == 0)
                return false;
        }
        return true;
    }

    static long gcd(long a, long b) {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }

    static void solve() {
        int n = ni();
        int k = ni();
        if (n <= (2 * k + 1)) {
            out.println(1 + "\n" + (n + 1) / 2);
        } else {
            int d = n / (2 * k + 1);
            int r = n % (2 * k + 1);
            if (r == 0) {
                out.println(d);
                int j = k + 1;
                while (j <= n) {
                    out.print(j + " ");
                    j += 2 * k + 1;
                }
            } else if (r >= k + 1) {
                out.println(d + 1);
                int j = k + 1;
                while (j <= n) {
                    out.print(j + " ");
                    j += 2 * k + 1;
                }
            } else {
                out.println(d + 1);
                int j = 1;
                while (j <= n) {
                    out.print(j + " ");
                    j += 2 * k + 1;
                }
            }
        }
    }
}