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本文介绍了一种通过算法解决学生考试座位安排的问题，确保没有两个相邻编号的学生坐在一起。使用C语言实现，针对不同学生数量提供了具体的解决方案。

采用插空法，注意特殊数据，1，2,3,4 尤其是3的情况输出应为 3 1 4 2败就败在这里了。其他情况均可以把1放在a[0]内，唯独3不可以。

#include<stdio.h>
#include<math.h>
#include<stdlib.h>

int main()
{
    int n;
    scanf("%d", &n);
    int a[5005] = {0};
    int b[5005] = {0};
    int i;
    int j;
       if( n == 4)
    {
        printf("4\n");
        printf("3 1 4 2");
    }
    else if( n == 3)
    {
        printf("2\n");
        printf("1 3");
    }
    else if( n == 2)
    {
        printf("1\n");
        printf("1");
    }
    else if(n == 1)
    {
        printf("1\n");
        printf("1");
    }
    else
    {
        for(i = 0, j = 1 ; j <= (n +1 )/ 2 ; i += 2, j++)
        {
            a[i] = j;
        }
        for(i = 1; j <= n; j++, i+=2 )
        {
            a[i] = j;
        }
        for(i = 0 ; a[i] != 0; i++);
        printf("%d\n", i);
        for(i = 0 ; a[i] != 0; i++)
        {
            printf("%d ",a[i]);
        }
    }

    return 0;
}

</pre><pre>

Description

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, ..., a_k (1 ≤ a_i ≤ n), where a_i 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: |a_i - a_i + 1| ≠ 1 for all i from 1 tok - 1.