Slava and tanks 877C

C. Slava and tanks

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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, ..., km. The number ki means that the i-th bomb should be dropped at the cell ki.

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

Examples

input

Copy

2

output

Copy

3
2 1 2 

input

Copy

3

output

Copy

4
2 1 3 2 
题意：炸坦克，如果炸第一个坦克，它只能跑到第二个坦克，如果炸第n个坦克，它只能跑到第n-1个坦克，如果炸中间坦克，它可以跑到它相邻的坦克上
分析：先炸偶数位上的坦克（因为偶数总是比奇数少），然后偶数位上的坦克都跑到奇数位上了，再炸奇数位上的坦克，原来偶数位上的坦克就没了，奇数位上的所有坦克跑到偶数位上去了，然后再炸一次偶数位上的坦克，这下所有的坦克都炸没了

 1 #include<cstdio>
 2 int main()
 3 {
 4     int n;
 5     while(~scanf("%d",&n))
 6     {
 7         printf("%d\n",3*n/2);//总次数
 8         for(int i=2;i<=n;i+=2)
 9         {
10             printf("%d ",i);
11         }
12         for(int i=1;i<=n;i+=2)
13         {
14             printf("%d ",i);
15         }
16         for(int i=2;i<=n;i+=2)
17         {
18             printf("%d ",i);
19         }
20         printf("\n");
21     }
22     return 0;
23 }