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.util.*;
import java.lang.*;
import java.lang.reflect.Array;
import java.io.*;
import java.math.*;
import java.text.DecimalFormat;
public class Prac{     
    static class InputReader { 
        private final InputStream stream;
        private final byte[] buf = new byte[8192];
        private int curChar, snumChars;
        public InputReader(InputStream st) {
            this.stream = st;
        } 
        public int read() {
            if (snumChars == -1)
                throw new InputMismatchException();
            if (curChar >= snumChars) {
                curChar = 0;
                try {
                    snumChars = stream.read(buf);
                } 
                catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (snumChars <= 0)
                    return -1;
            }
            return buf[curChar++];
        }
        public int ni() {
            int c = read();
            while (isSpaceChar(c)) {
                c = read();
            }
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = read();
            }
            int res = 0;
            do {
                res *= 10;
                res += c - '0';
                c = read();
            } while (!isSpaceChar(c));
            return res * sgn;
        } 
        public long nl() {
            int c = read();
            while (isSpaceChar(c)) {
            c = read();
            }
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = read();
            }
            long res = 0;
            do {
                res *= 10;
                res += c - '0';
                c = read();
            } while (!isSpaceChar(c));
            return res * sgn;
        } 
        public int[] nia(int n) {
            int a[] = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = ni();
            }
            return a;
        } 
        public String rs() {
            int c = read();
            while (isSpaceChar(c)) {
                c = read();
            }
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                    c = read();
            } while (!isSpaceChar(c));
            return res.toString();
        }
        public String nextLine() {
            int c = read();
            while (isSpaceChar(c))
                c = read();
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                c = read();
            } while (!isEndOfLine(c));
            return res.toString();
        } 
        public boolean isSpaceChar(int c) {
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }

        private boolean isEndOfLine(int c) {
            return c == '\n' || c == '\r' || c == -1;
        } 
    }
    static PrintWriter w = new PrintWriter(System.out);
    static long mod=998244353L,mod1=1000000007;

    public static class Key {

    private final int x;
    private final int y;
    private final int z;
    public Key(int x, int y,int z) {
        this.x = x;
        this.y = y;
        this.z=z;
    }

    @Override
    public boolean equals(Object o) {
        if (this == o) return true;
        if (!(o instanceof Key)) return false;
        Key key = (Key) o;
        return x == key.x && y == key.y&&z==key.z;
    }

    @Override
    public int hashCode() {
        int result = x;
        result = 31 * result + y;
        result=31*result+z;
        return result;
    }

}
    
    public static void main (String[] args)throws IOException{
        InputReader sc=new InputReader(System.in);
        int n=sc.ni();
        int k=sc.ni();
        int dp[]=new int[n];
        Arrays.fill(dp,Integer.MAX_VALUE);
        int i=k+1;
        ArrayList<Integer> ans=new ArrayList<>();
        boolean f=false;
        while(i<=n){
            ans.add(i);
//            if(n-(i+2*k+1)<0&&(i+k)<=n){
//                f=true;
//                break;
//            }
//            else {
                 i=i+2*k+1;
            //}
        }
        if(ans.size()<Math.ceil((double)n/(double)(2*k+1))){
            ans.clear();
            i=1;
            while(i<=n){
                ans.add(i);
                i=i+2*k+1;
            }
        }
        w.println(ans.size());
        for(int j:ans)w.print(j+" ");
        w.println();
        w.close();
    }
}