LCM Challenge--codeForces 235A

Description

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 thann. Can you help me to find the maximum possible least common multiple of these three integers?

Input

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

Output

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

Sample Input

Input

9

Output

504

Input

7

Output

210

首先考虑n是奇数的情况：显然n和n-1互质，n-1和n-2也是互质的，n和n-2呢。相差2.唯一可能让他们不互质的因子只能是2.他们都是奇数。所以，这3个数的最小公倍数是n*(n-1)*(n-2)。绝对最大

然后是偶数的情况：偶数的话，(n-1)*(n-2)*(n-3)依然是个很大的最小公倍数，我们只需要找找看有没有比他大的就行了。唯一可能就是n和n-1,然后再找一个。注意，当n>=6的时候，n,n-1,n-2的最小公倍数是比n-1,n-2,n-3的最小公倍数小的。循环查找即可。

#include 
#include 
using namespace std;
long long int gcd(long long int a,long long int b)
{
	if(b==0)return a;
	return gcd(b,a%b);
}
long long int lcm(long long int a,long long int b)
{
	long long int c=gcd(a,b);
	return a*b/c;
}
int main()
{
	long long int n;
	while(scanf("%I64d",&n)!=EOF)
	{
		if(n==1)cout<<1<=6)
		{
			if(n%2)
			{
				cout<fuck;i-=2)
				{
					if(lcm(fuckyou,i)>fuck)
					{
						fuck=lcm(fuckyou,i);
						break;
					}
				}
				cout<