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:

/**
 * Date: 5 Sep, 2018
 * Link:
 *
 * @author Prasad-Chaudhari
 * @linkedIn: https://www.linkedin.com/in/prasad-chaudhari-841655a6/
 * @git: https://github.com/Prasad-Chaudhari
 */
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.LinkedList;

public class newProgram4 {

    public static void main(String[] args) throws IOException {
        // TODO code application logic here
        FastIO in = new FastIO();
        int n = in.ni();
        int k = in.ni();
        LinkedList<Integer> l = new LinkedList<>();
        int start = 1;
        while (start <= n) {
            l.add(start);
            start += 2 * k + 1;
        }
        int diff = -1;
        if (l.getLast() + k <= n) {
            diff = n - l.getLast() - k;
        }
        System.out.println(l.size());
        for (int i : l) {
            if (diff != -1) {
                System.out.println(i + diff);
            }
            else{
                System.out.println(i);
            }
        }
    }

    static class FastIO {

        private final BufferedReader br;
        private final BufferedWriter bw;
        private String s[];
        private int index;
        private StringBuilder sb;

        public FastIO() throws IOException {
            br = new BufferedReader(new InputStreamReader(System.in));
            bw = new BufferedWriter(new OutputStreamWriter(System.out, "UTF-8"));
            s = br.readLine().split(" ");
            sb = new StringBuilder();
            index = 0;
        }

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

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

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

        public String next() throws IOException {
            return nextToken();
        }

        public String nli() throws IOException {
            try {
                return br.readLine();
            } catch (IOException ex) {

            }
            return null;
        }

        public int[] gia(int n) throws IOException {
            int a[] = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = ni();
            }
            return a;
        }

        public int[] gia(int n, int start, int end) throws IOException {
            validate(n, start, end);
            int a[] = new int[n];
            for (int i = start; i < end; i++) {
                a[i] = ni();
            }
            return a;
        }

        public double[] gda(int n) throws IOException {
            double a[] = new double[n];
            for (int i = 0; i < n; i++) {
                a[i] = nd();
            }
            return a;
        }

        public double[] gda(int n, int start, int end) throws IOException {
            validate(n, start, end);
            double a[] = new double[n];
            for (int i = start; i < end; i++) {
                a[i] = nd();
            }
            return a;
        }

        public long[] gla(int n) throws IOException {
            long a[] = new long[n];
            for (int i = 0; i < n; i++) {
                a[i] = ni();
            }
            return a;
        }

        public long[] gla(int n, int start, int end) throws IOException {
            validate(n, start, end);
            long a[] = new long[n];
            for (int i = start; i < end; i++) {
                a[i] = ni();
            }
            return a;
        }

        public int[][] gg(int n, int m) throws IOException {
            int adja[][] = new int[n + 1][];
            int from[] = new int[m];
            int to[] = new int[m];
            int count[] = new int[n + 1];
            for (int i = 0; i < m; i++) {
                from[i] = ni();
                to[i] = ni();
                count[from[i]]++;
                count[to[i]]++;
            }
            for (int i = 0; i <= n; i++) {
                adja[i] = new int[count[i]];
            }
            for (int i = 0; i < m; i++) {
                adja[from[i]][--count[from[i]]] = to[i];
                adja[to[i]][--count[to[i]]] = from[i];
            }
            return adja;
        }

        public void print(String s) throws IOException {
            bw.write(s);
            bw.flush();
        }

        public void println(String s) throws IOException {
            bw.write(s);
            bw.newLine();
            bw.flush();
        }

        private String nextToken() throws IndexOutOfBoundsException, IOException {
            if (index == s.length) {
                s = br.readLine().split(" ");
                index = 0;
            }
            return s[index++];
        }

        private void validate(int n, int start, int end) {
            if (start < 0 || end >= n) {
                throw new IllegalArgumentException();
            }
        }
    }

    static class Data implements Comparable<Data> {

        int a, b;

        public Data(int a, int b) {
            this.a = a;
            this.b = b;
        }

        @Override
        public int compareTo(Data o) {
            return Integer.compare(a, o.a);
        }
    }
}