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.BufferedReader;
import java.util.*;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.io.PrintWriter;
public class faltu {
	int INF = (int) 1e9;
	int MOD = 1000000007;
    byte broken[][];
    
    long pow(int a ,int b){
        if(b==0)
            return 1;
        if(b==1)
            return a;
        if(b%2==0){
            long x = pow(a,b/2);
            return x*x%MOD;
        }
        else{
            return pow(a,b-1)*a%MOD;
        }
    }
    
    ArrayList<ArrayList<Integer>> g = new ArrayList<>();
    boolean visited[];
    boolean recStack[];
    int count=0;
    int deg[];
    
	void solve(InputReader in, PrintWriter out) throws IOException {
		int n = in.nextInt();
        int k = in.nextInt();
        int x = 2*k+1;
        int y = (int)Math.ceil(n*1.0/x);
        
        int p=k+1;
        if(n%(x)>k || n%x==0){
            System.out.println(y);
            for(int i=0;i<y;i++){
                    System.out.print(p+" ");
                    p+=x;
            }
        }
        else{
            p=1;
            System.out.println(y);
            for(int i=0;i<y;i++){
                    System.out.print(p+" ");
                    p+=x;
            }
        }
	}

	public static void main(String[] args) throws IOException {
		InputReader in = new InputReader();
		PrintWriter out = new PrintWriter(System.out);
		int t = 1;//in.nextInt();
//		long arr[] = new long[1000000];
//		for(int i=0;i<1000000;i++){
//			arr[i]=(i+1)*(i+2)/2;
//		}
		while(t-- > 0) {
			new faltu().solve(in, out);
		}
		out.close();
	}

	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());
		}
	}
}