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:
/*
[ C O D E    T I L L    I N F I N I T Y ]
*/
import java.io.*;
import java.util.*;


public class test {
    int INF = (int)1e9;
    long MOD = 1000000007;
    
    
    void solve(InputReader in, PrintWriter out) throws IOException {
        int n = in.nextInt();
        int k = in.nextInt();
        int r = 2 * k + 1;
        if(r>=n) {
            int p = (n+1)/2;
            out.println(1);
            out.println(p);
            return;
        }
        int m = n % r;
        int p = k+1;
        if(m>0) {
            p = Math.min(m, k+1);
        }
        ArrayList<Integer> rs = new ArrayList<>();
        while(p<=n) {
            rs.add(p);
            p += r;
        }
        out.println(rs.size());
        for(int ri: rs) {
            out.print(ri+" ");
        }
        out.println();
    }
    
    
    public static void main(String[] args) throws IOException {
        if(args.length>0 && args[0].equalsIgnoreCase("d")) {
            DEBUG_FLAG = true;
        }
        InputReader in = new InputReader();
        PrintWriter out = new PrintWriter(System.out);
        int t = 1;//in.nextInt();
        long start = System.nanoTime();
        while(t-->0) {
            new test().solve(in, out);
        }
        long end = System.nanoTime();
        debug("\nTime: " + (end-start)/1e6 + " \n\n");
        out.close();
    }
    
    static boolean DEBUG_FLAG = false;
    static void debug(String s) {
        if(DEBUG_FLAG) System.out.print(s);
    }
    
    public static void show(Object... o) {
        System.out.println(Arrays.deepToString(o));
    }
    
    static class InputReader {
        static BufferedReader br;
        static StringTokenizer st;
    
        public InputReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }
        
        String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        
        int nextInt() {
            return Integer.parseInt(next());
        }
        
        long nextLong() {
            return Long.parseLong(next());
        }
        
        double nextDouble() {
            return Double.parseDouble(next());
        }
    }
    
    class Pair implements Comparable<Pair> {
        long f, s;
        
        Pair(long f, long s) {
            this.f=f;
            this.s=s;
        }
        
        public int hashCode() {
            int hf = (int) (f ^ (f >>> 32));
            int hs = (int) (s ^ (s >>> 32));
            return 31 * hf + hs;
        }
        
        public boolean equals(Object o) {
            Pair other = (Pair) o;
            return s == other.s && f == other.f;
        }
        
        public int compareTo(Pair p) {
            return Long.compare(this.s, p.s);
        }
        
        public String toString() {
            return "(" + f + ", " + s + ")";
        }
    }
}