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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.*;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.FileOutputStream;
import java.io.InputStreamReader;
import java.io.InputStream;

public class Main {
	
    public static void main(String[] args) throws FileNotFoundException {
//    	File file1 = new File("C:/Users/hasee/Desktop/std.in");
//    	File file3 = new File("C:/Users/hasee/Desktop/std.out");
//    	File file2 = new File("C:/Users/hasee/Desktop/2.txt");
//    	InputStream inputStream = new FileInputStream(file1);
//    	InputStream inputStream1 = new FileInputStream(file3);
//    	OutputStream outputStream = new FileOutputStream(file2);
    	InputStream inputStream = System.in;
    	OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskA solver = new TaskA();
        solver.solve(1, in, out);
        out.close();
    }
    
    static class TaskA {
    	
        public void solve(int testNumber, InputReader in, PrintWriter out) {    
        	
        	int a = in.nextInt();
        	int b = in.nextInt();
        	if(a<=b*2+1){
        		System.out.println(1);
        		System.out.println(a/2+1);
        	}
        	else if(a%(b*2+1)==0){
        		System.out.println(a/(b*2+1));
        		int t = b+1;
        		for(int n=0;n<a/(b*2+1);n++){
        			System.out.println(t);
        			t += (2*b+1);
        		}
        	}
        	else{
        		
        		int c = a%(b*2+1);
        		if(c >= b+1){
        			System.out.println(a/(b*2+1)+1);
        			int t = c - b;
        			for(int n=0;n<a/(b*2+1)+1;n++){
        				if(n!=0)System.out.print(" ");
        				System.out.print(t);
        				
            			t += (2*b+1);
        			}
        		}
        		else{
        			System.out.println(a/(b*2+1)+1);
        			int t = 1;
        			for(int n=0;n<a/(b*2+1)+1;n++){
        				if(n!=0)System.out.print(" ");
        				System.out.print(t);
            			t += (2*b+1);
        			}
        		}
        		
        	}
        	
        	
        }
        
    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }
        
        public double nextDouble() {
        	return Double.parseDouble(next());
        }
        
        public long nextLong() {
            return Long.parseLong(next());
        }
        
    }
}

class Node{
	int l,r,t;
}