Problem: An exam for n students will take place in a long and narrow room, so the students will sit in a line in some order. The teacher suspects that students with adjacent numbers (i and i + 1) always studied side by side and became friends and if they take an exam sitting next to each other, they will help each other for sure.

Your task is to choose the maximum number of students and make such an arrangement of students in the room that no two students with adjacent numbers sit side by side.

Input

A single line contains integer n (1 ≤ n ≤ 5000) — the number of students at an exam.

Output

In the first line print integer k — the maximum number of students who can be seated so that no two students with adjacent numbers sit next to each other.

In the second line print k distinct integers a1, a2, ..., ak (1 ≤ ai ≤ n), where ai is the number of the student on the i-th position. The students on adjacent positions mustn't have adjacent numbers. Formally, the following should be true: |ai - ai + 1| ≠ 1 for all i from 1 to k - 1.

If there are several possible answers, output any of them.

Examples

Input

6

Output

6
1 5 3 6 2 4

Input

3


Output

2
1 3

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.StringTokenizer;

/**
 * Created by peacefrog on 11/8/15.
 * Time : 12:24 PM
 */

public class Exam289A {
	final boolean ONLINE_JUDGE = System.getProperty("ONLINE_JUDGE") != null;
	PrintWriter out;
	long timeBegin, timeEnd;

	public void runIO() throws IOException {
		timeBegin = System.currentTimeMillis();

		InputStream inputStream;
		OutputStream outputStream;


		if (ONLINE_JUDGE) {
			inputStream = System.in;
			Reader.init(inputStream);
			outputStream = System.out;

			out = new PrintWriter(outputStream);

		} else {
			inputStream = new FileInputStream("/home/peacefrog/Dropbox/IdeaProjects/Problem Solving/input");
			Reader.init(inputStream);
			out = new PrintWriter("output.txt");
		}

		solve();

		out.flush();
		out.close();
		timeEnd = System.currentTimeMillis();

		System.err.println("Time = " + (timeEnd - timeBegin));

	}

	/*
	 * Start Solution Here
	 */

	private void solve() throws IOException {
		int TT = Reader.nextInt() ;
		if (TT <= 2 ) out.println("1\n1");
		else if (TT == 3 ) out.println("2\n1 3");
		else if (TT == 4 ) out.println("4\n3 1 4 2");
		else {

			out.println(TT);
			for (int i = 1; i <= TT; i++) {
				if ((i & 1)==1) out.print(i  + " ");
			}
			for (int i = 1; i <= TT; i++) {
				if ((i & 1)==0) out.print(i  + " ");
			}
		}

	}

	public static void main(String[] args) throws IOException {
		new Exam289A().runIO();
	}

	static class Reader {
		static BufferedReader reader;
		static StringTokenizer tokenizer;

		/**
		 * call this method to initialize reader for InputStream
		 */
		static void init(InputStream input) {
			reader = new BufferedReader(new InputStreamReader(input));
			tokenizer = new StringTokenizer("");
		}

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

		static String nextLine() {
			try {
				return reader.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return "";
		}

		static int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		static long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		static double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}
	}
}