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:
//Author: Mo Abjal, MJP Rohilkhand University Bareilly, UP, India 2018.
// NSR - E-Card
import java.util.*;
import java.io.*;
import java.text.*;
import java.math.*;
public class Main{
	public static StringTokenizer token = new StringTokenizer("");
	public static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	public static PrintWriter output = new PrintWriter(System.out);
	public static void main(String[] afzal) throws Exception {		
		int N = ni();
		int K = ni();
		if(K==0){
			output.println(N);
			for(int i=1;i<=N;i++) output.print(i+" ");
		}
		else if(N<=K){
			output.println("1\n1");
		}
		else{
			//73000861291
			ArrayList<Integer> list = new ArrayList<Integer>();
			int step = K+1;
			int min = 2*K+1 - N%(2*K+1);
			if(min==2*K+1){
				min = 0;
			}
			if(min>K) min=K;
			for(int i=step-min;i<=N;i=i+2*K+1){
				list.add(i);
			}
			output.println(list.size());
			for(int i:list) output.print(i+" ");
			
		}
		output.close();
	}
	//Scanning Inputs and Converting data library
	//Parse Integer
	public static Integer pi(String str){
		return Integer.parseInt(str);
	}
	//Parse Long
	public static Long pl(String str){
		return Long.parseLong(str);
	}
	//Converting Integer into String
	public static String ps(Integer N){
		return Integer.toString(N);
	}
	//converting string into char Array
	public static char[] pc(String str){
		return str.toCharArray();
	}
	//Taking Integer Input
	public static Integer ni() throws IOException{
		if(!token.hasMoreElements()){
			token = new StringTokenizer(br.readLine());
		}
		return Integer.parseInt(token.nextToken());
	}
	//Taking Long Input
	public static Long nl() throws IOException{
		if(!token.hasMoreElements()){
			token = new StringTokenizer(br.readLine());
		}
		return Long.parseLong(token.nextToken());
	}
	//Taking String Input
	public static String ns() throws IOException{
		return new StringTokenizer(br.readLine()).nextToken();
	}
	//Taking Integer Array
	public static Integer[] ai(int N) throws IOException{
		Integer A[] = new Integer[N];
		token = new StringTokenizer(br.readLine());
		for(int i=0;i<A.length;i++){
			A[i] = pi(token.nextToken());
		}
		return A;
	}
	//Taking Long Array
	public static long[] al(int N) throws IOException{
		long A[] = new long[N];
		token = new StringTokenizer(br.readLine());
		for(int i=0;i<N;i++){
			A[i] = Long.parseLong(token.nextToken());
		}
		return A;
	}
	//Taking 2D Char Array
	public static char[][] acc(int N,int M) throws IOException{
		char C[][] = new char[N][M];
		for(int i=0;i<N;i++){
			token = new StringTokenizer(br.readLine());
			String s = token.nextToken();
			for(int j=0;j<M;j++){
				C[i][j] = s.charAt(j);
			}
		}
		return C;
	}
	//Taking String Array
	public static String[] as(int N) throws IOException{
		String S[] = new String[N];
		token = new StringTokenizer(br.readLine());
		for(int i=0;i<S.length;i++){
			S[i] = token.nextToken();
		}
		return S;
	}

	//Calculating factorial
	public static long fact(long N){
		long fact = 1;
		for(int i=1;i<=N;i++){
			fact = fact * i;
		}
		return fact;
	}
	public static BigInteger factB(int N){
		BigInteger b = new BigInteger("1");
		for(int i=1;i<=N;i++){
			b = b.multiply(BigInteger.valueOf(i));
		}
		return b;
	}
	//Making Paares of given size
	public static long mp(long N,long size){
		long mul=1;
		long temp = N;
		for(long i=N;i>N-size;i--){
			mul = mul*temp--;
		}
		long fact = fact(size);
		if(mul/fact==0) return 1;
		return mul/fact;
	}
	public static long sod(long N){
		long sum = 0;
		while(N>0){
			long x = N%10;
			sum = sum + x;
			N = N/10;
		}
		return sum;
	}
}



//Copyright Reserved - Mo Abjal.