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?

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.

Examples

Input

9

Output

504

Input

7

Output

210

Note

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.

 

题意：输出由三个不大于n的数所组成的最大的最小公倍数

n为奇数，首先要明白 奇数*奇数=奇数，显然此时n*n-1*n-2(此时由于n为奇数，所以n与n-2不存在任何交集)最大

n为偶数时，n*n-1*n-3 看似是正确答案，但由于 n 与 n-3 可能同时被3整除，所以进一步判断，若能被3整除，则

ans=n-1*n-2*n-3

//#pragma GCC optimize(2)
#include <bits/stdc++.h>
#define rush() int T;cin>>T;while(T--)
#define go(a) while(cin>>a)
#define ms(a,b) memset(a,b,sizeof a)
#define E 1e-8
#define debug(a) cout<<"*"<<a<<"*"<<endl
#define IOS ios::sync_with_stdio(0); cin.tie(0);// cout.tie(0)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int inf=0x3f3f3f3f;
const int N=1000+5;

    int n,m,t;
    int i,j,k;

int32_t main()
{
    IOS;
    while(cin>>n){
        if(n==1||n==2) {cout<<n<<endl;continue;}
        if(n&1) cout<<1LL*n*(n-1)*(n-2)<<endl;
        else{
            if(n%3==0) cout<<1LL*(n-2)*(n-1)*(n-3)<<endl;
            else cout<<1LL*n*(n-1)*(n-3)<<endl;
        }
    }
    return 0;
}