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.*;
import java.util.*;
public class Main
{
	public static void main(String[] args) throws Exception
	{
		// int t = ni();
		int t = 1;
		while(t-->0) {
			int n = ni();
			int k = ni();
			int blocks = k*2 + 1;
			int min = k+1;
			if(n%blocks == 0){
				sopln(n/blocks);
			}
			else
				sopln(n/blocks + 1);
			if(n%blocks >= min || n%blocks == 0){
				for(int i=min;i<=n;i+=blocks){
					ans.append(i+" ");
				}
			}
			else {
				int rem = blocks - n%blocks;
				for(int i=min - (rem/2);i<=n;i+=blocks){
					ans.append(i+" ");
				}
			}
		}
		sop(ans);
	}
	static final long mod=1000000007l;
    static StringBuilder ans=new StringBuilder();
	static Reader in=new Reader();
	static final double eps=1e-9;
 
	//////////////////////////////////
	////input functions/////////////////
	static int ni(){return Integer.parseInt(in.next());}
	static long nl(){return Long.parseLong(in.next());}
	static String ns(){return in.next();}
	static double nd(){return Double.parseDouble(in.next());}
	static int[] nia(int n){int a[]=new int[n];for(int i=0; i<n; i++)a[i]=ni();return a;}
	static int[] pnia(int n){int a[]=new int[n+1];for(int i=1; i<=n; i++)a[i]=ni();return a;}
	static long[] nla(int n){long a[]=new long[n];for(int i=0; i<n; i++)a[i]=nl();return a;}
	static String[] nsa(int n){String a[]=new String[n];for(int i=0; i<n; i++)a[i]=ns();return a;}
	static double[] nda(int n){double a[]=new double[n];for(int i=0; i<n; i++)a[i]=nd();return a;}
 
	//////output functions////////////////
	static void pr(Object a){ans.append(a+"\n");}
	static void pr(){ans.append("\n");}
	static void p(Object a){ans.append(a);}
	static void pra(int[]a){for(int i:a)ans.append(i+" ");ans.append("\n");}
	static void pra(long[]a){for(long i:a)ans.append(i+" ");ans.append("\n");}
	static void pra(String[]a){for(String i:a)ans.append(i+" ");ans.append("\n");}
	static void pra(double[]a){for(double i:a)ans.append(i+" ");ans.append("\n");}
	static void sop(Object a){System.out.print(a);}
	static void sopln(Object a){System.out.println(a);}
	static void flush(){System.out.print(ans);ans=new StringBuilder();}
 
	static void exit(){System.out.print(ans);System.exit(0);}
	static int min(int... a){int min=a[0];for(int i:a)min=Math.min(min, i);return min;}
	static int max(int... a){int max=a[0];for(int i:a)max=Math.max(max, i);return max;}	
	static int gcd(int... a){int gcd=a[0];for(int i:a)gcd=gcd(gcd, i);return gcd;}	
	static long min(long... a){long min=a[0];for(long i:a)min=Math.min(min, i);return min;}
	static long max(long... a){long max=a[0];for(long i:a)max=Math.max(max, i);return max;}	
	static long gcd(long... a){long gcd=a[0];for(long i:a)gcd=gcd(gcd, i);return gcd;}	
	static String pr(String a, long b){String c="";while(b>0){if(b%2==1)c=c.concat(a);a=a.concat(a);b>>=1;}return c;}
	static long powm(long a, long b, long m){long an=1;long c=a;while(b>0){if(b%2==1)an=(an*c)%m;c=(c*c)%m;b>>=1;}return an;}
	static int gcd(int a, int b){if(b==0)return a;return gcd(b, a%b);}
	static long gcd(long a, long b){if(b==0)return a;return gcd(b, a%b);}
 
	static class Reader {
        BufferedReader reader;
        StringTokenizer tokenizer;
        public Reader() {
            reader = new BufferedReader(new InputStreamReader(System.in), 32768);
            tokenizer = null;
        }
        String next() {while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {tokenizer = new StringTokenizer(reader.readLine());}
				catch (IOException e) {throw new RuntimeException(e);}}
            return tokenizer.nextToken();}
        int nextInt() {return Integer.parseInt(next());}
        long nextLong() {return Long.parseLong(next());}
        double nextDouble() {return Double.parseDouble(next());}
    }
 
}