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Slava plays his favorite game "Peace Lightning". Now he is flying a bomber on a very specific map.

Formally, map is a checkered field of size 1 × n, the cells of which are numbered from 1 to n, in each cell there can be one or several tanks. Slava doesn't know the number of tanks and their positions, because he flies very high, but he can drop a bomb in any cell. All tanks in this cell will be damaged.

If a tank takes damage for the first time, it instantly moves to one of the neighboring cells (a tank in the cell n can only move to the cell n - 1, a tank in the cell 1 can only move to the cell 2). If a tank takes damage for the second time, it's counted as destroyed and never moves again. The tanks move only when they are damaged for the first time, they do not move by themselves.

Help Slava to destroy all tanks using as few bombs as possible.

Input

The first line contains a single integer n (2 ≤ n ≤ 100 000) — the size of the map.

Output

In the first line print m — the minimum number of bombs Slava needs to destroy all tanks.

In the second line print m integers k₁, k₂, ..., k_m. The number k_i means that the i-th bomb should be dropped at the cell k_i.

If there are multiple answers, you can print any of them.

Example

Input

2

Output

32 1 2 

Input

3

Output

42 1 3 2 

题意： 输入一个数n，n代表有n个位置，每个位置可能有坦克，可能也没有坦克，一个坦克要炸两次才能毁掉，若这个位置有坦克，在这个位置炸一下，这个位置的坦克就会跑到左右相邻的其中一个位置，要把所有坦克都炸毁，求最小的轰炸数；

思路：因为在这个位置炸一下，这个位置的坦克就会跑到左右相邻的其中一个位置，你不知道会跑左边还是右边，所以你就分奇偶，炸奇数的坦克一定会跑偶数里去，炸偶数的坦克一定会跑奇数里去，所以分奇偶数炸，那么是先炸奇数呢，还是先炸偶数呢？先炸偶数，因为给出一个数n，在从1~n的整数中，偶数的数量一定会小于等于奇数的数量，

所以先炸奇数，再炸偶数，再炸奇数；

代码：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int main(){int i,j,n;while(~scanf("%d",&n)){printf("%d\n",n+n/2);if(n==1){printf("1 1\n");continue;}printf("2");for(i=4;i<=n;i+=2)printf(" %d",i);for(i=1;i<=n;i+=2)printf(" %d",i);for(i=2;i<=n;i+=2)printf(" %d",i);printf("\n");}return 0;}