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:
//package com.company;

import java.io.*;
import java.util.*;

public class Main {
    static long TIME_START, TIME_END;
    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
//        Scanner sc = new Scanner(new FileInputStream("Test.in"));
        PrintWriter pw = new PrintWriter(System.out);
//        PrintWriter pw = new PrintWriter(new FileOutputStream("Test.out"));
//        PrintWriter pw = new PrintWriter(new FileOutputStream("Test.in"));

        Runtime runtime = Runtime.getRuntime();
        long usedMemoryBefore = runtime.totalMemory() - runtime.freeMemory();
        TIME_START = System.currentTimeMillis();
        Task t = new Task();
        t.solve(sc, pw);
        TIME_END = System.currentTimeMillis();
        long usedMemoryAfter = runtime.totalMemory() - runtime.freeMemory();
        pw.close();
//        System.out.println("Memory increased:" + (usedMemoryAfter-usedMemoryBefore) / 1000000 );
//        System.out.println("Time used: " + (TIME_END - TIME_START) + ".");


    }

    public static class Task {

        public void solve(Scanner sc, PrintWriter pw) throws IOException {
            int n = sc.nextInt();
            if (n == 1) {
                pw.println(1);
                pw.println(1);
                return;
            } else if (n == 2){
                pw.println(1);
                pw.println(1);
                return;
            } else if (n == 3) {
                pw.println(2);
                pw.println("1 3");
            } else if (n == 4){
                pw.println(4);
                pw.println("3 1 4 2");
            } else {
                pw.println(n);
                int[] arr = new int[n];
                for (int i = 0; i < (n + 1) / 2; i++) {
                    arr[i] = i * 2 + 1;
                }
                for (int i = (n + 1) / 2; i < n; i++) {
                    arr[i] = (i - (n + 1) / 2 + 1) * 2;
                }
                for (int i = 0; i < n; i++) {
                    pw.print(arr[i] + " ");
                }
            }

        }

    }

    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream s){  br = new BufferedReader(new InputStreamReader(s));}

        public Scanner(FileReader s) throws FileNotFoundException {br = new BufferedReader(s);}

        public String next() throws IOException
        {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

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

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

        public String nextLine() throws IOException {return br.readLine();}

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

        public boolean ready() throws IOException {return br.ready();}
    }
}