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.math.*;
import java.util.*;

import static java.util.Arrays.fill;
import static java.lang.Math.*;
import static java.util.Arrays.sort;
import static java.util.Collections.sort;

public class B {
	static int mod = 1000000007;
	static InputReader in = new InputReader(System.in);
	static PrintWriter out = new PrintWriter(System.out);

	public static void main(String[] args) {

		int n=in.nextInt();
		int k=in.nextInt();
		int[] a=new int[n+1];
		int mini=10000;
		int minpos=-1;
		for(int pos=1;pos<=n;pos++)
		{
			Arrays.fill(a, 0);
			int temp=0;
			temp++;
			a[pos]++;
			for(int i=1;i<=k;i++)
			{
				if(pos-i>=1)
				a[pos-i]++;
				if(pos+i<=n)
				a[pos+i]++;
			}
			for(int j=pos+2*k+1;j<=n;j+=(2*k+1))
			{
				temp++;
				a[j]++;
				for(int i=1;i<=k;i++)
				{
					if(j-i>=1)
					a[j-i]++;
					if(j+i<=n)
					a[j+i]++;
				}
			}
			boolean valid=true;
			for(int i=1;i<=n;i++)
			{
				if(a[i]%2==0)
				{
					valid=false;
					break;
				}
			}
			if(valid)
			{
				if(temp<mini)
				{
					mini=temp;
					minpos=pos;
				}
			}
		}
		out.println(mini);
		out.print(minpos+" ");
		for(int i=minpos+2*k+1;i<=n;i+=(2*k+1))
		{
			out.print(i+" ");
		}
		out.close();

	}

	static class InputReader {
		private final InputStream stream;
		private final byte[] buf = new byte[8192];
		private int curChar, snumChars;

		public InputReader(InputStream st) {
			this.stream = st;
		}

		public int read() {
			if (snumChars == -1)
				throw new InputMismatchException();
			if (curChar >= snumChars) {
				curChar = 0;
				try {
					snumChars = stream.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (snumChars <= 0)
					return -1;
			}
			return buf[curChar++];
		}

		public int nextInt() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			int res = 0;
			do {
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}

		public long nextLong() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			long res = 0;
			do {
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}

		public String readString() {
			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 String nextLine() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isEndOfLine(c));
			return res.toString();
		}

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

		private boolean isEndOfLine(int c) {
			return c == '\n' || c == '\r' || c == -1;
		}
	}

	private static void tr(Object... o) {
		if (!(System.getProperty("ONLINE_JUDGE") != null))
			System.out.println(Arrays.deepToString(o));
	}
}