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.awt.List;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.text.DecimalFormat;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.InputMismatchException;
import java.util.Locale;
import java.util.TimeZone;

public class B {

	public static void main(String[] args) {
		Locale.setDefault(Locale.US);
		InputStream inputstream = System.in;
		OutputStream outputstream = System.out;
		FastReader in = new FastReader(inputstream);
		PrintWriter out = new PrintWriter(outputstream);
		TaskA solver = new TaskA();
		// int testcase = in.ni();
		for (int i = 0; i < 1; i++)
			solver.solve(i, in, out);
		out.close();

	}

}

class TaskA {

	public void solve(int testnumber, FastReader in, PrintWriter out) {
		int n = in.ni();
		int k = in.ni();
		
		int ans = n / (2 * k + 1);
		int rem = n % (2 * k + 1);
		if(rem > 0 ){
			ans++;
		}
		out.println(ans);
		
		if(rem > k || rem == 0){
			int pos = k+1;
			ArrayList<Integer> arr = new ArrayList<>();
			while(pos<=n){
				arr.add(pos);
				pos += 2 * k + 1;
			}
			boolean flag = true;
			for(int value : arr){
				if(flag){
					flag = false;
				}else{
					out.print(" ");
				}
				out.print(value);
			}
			out.println();
		}else{
			int pos = 1;
			ArrayList<Integer> arr = new ArrayList<>();
			while(pos<=n){
				arr.add(pos);
				pos += 2 * k + 1;
			}
			boolean flag = true;
			for(int value : arr){
				if(flag){
					flag = false;
				}else{
					out.print(" ");
				}
				out.print(value);
			}
			out.println();
		}
	}
}

class FastReader {
	public InputStream stream;
	private byte[] buf = new byte[1024];
	private int curChar;
	private int numChars;
	private SpaceCharFilter filter;

	public FastReader(InputStream stream) {
		this.stream = stream;
	}

	public FastReader() {

	}

	public int read() {
		if (numChars == -1) {
			throw new InputMismatchException();
		}
		if (curChar >= numChars) {
			curChar = 0;
			try {
				numChars = stream.read(buf);
			} catch (IOException e) {
				throw new InputMismatchException();
			}
			if (numChars <= 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 {
			if (c < '0' || c > '9') {
				throw new InputMismatchException();
			}
			res *= 10;
			res += c - '0';
			c = read();
		} while (!isSpaceChar(c));
		return res * sgn;
	}

	public String ns() {
		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 boolean isSpaceChar(int c) {
		if (filter != null) {
			return filter.isSpaceChar(c);
		}
		return isWhitespace(c);
	}

	public static boolean isWhitespace(int c) {
		return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
	}

	public int[] iArr(int n) {
		int a[] = new int[n];
		for (int i = 0; i < n; i++) {
			a[i] = ni();
		}
		return a;
	}

	public String next() {
		return ns();
	}

	public interface SpaceCharFilter {
		public boolean isSpaceChar(int ch);
	}

}