CF235A.LCM Challenge

Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I want to make a big number with it.

But I also don't want to use many numbers, so I'll choose three positive integers (they don't have to be distinct) which are not greater than n. Can you help me to find the maximum possible least common multiple of these three integers?

几天前，我学习了最小公倍数（LCM）的概念。我已多次尝试用它构造数字，现在我想用它来构造一个尽可能大的数。

但同时，我又不想使用太多数字，因此我将从不超过 $n$ 的正整数中选出三个数（它们可以相同）。你能帮我找出这三个整数可能达到的最大最小公倍数吗？

The first line contains an integer n (1 ≤ n ≤ 106) — the n mentioned in the statement.

第一行包含一个整数 $n$（$1 \leq n \leq 10^6$）——即题目描述中提到的 $n$。

Print a single integer — the maximum possible LCM of three not necessarily distinct positive integers that are not greater than n.

输出一个整数——不超过 $n$ 的三个（不一定互不相同）正整数所能达到的最大可能的最小公倍数（LCM）。

The least common multiple of some positive integers is the least positive integer which is multiple for each of them.

The result may become very large, 32-bit integer won't be enough. So using 64-bit integers is recommended.

For the last example, we can chose numbers 7, 6, 5 and the LCM of them is 7·6·5 = 210. It is the maximum value we can get.

若干正整数的最小公倍数（LCM）是指能被其中每一个数整除的最小正整数。

结果可能非常大，32 位整数不足以表示，因此建议使用 64 位整数。

对于最后一个例子，我们可以选择数字 $7$、$6$、$5$，它们的最小公倍数为 $7 \cdot 6 \cdot 5 = 210$。这是所能得到的最大值。