#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<string>
#include<cstring>
#include<set>
#include<cmath>
#include<queue>
#define int long long
#define ldou long double
#define endl '\n'
#define F first
#define S second
#define sz size()
#define B begin()
#define D end()
#define pii pair<int,int>
#define IOS ios::sync_with_stdio(false)
#define lowbit(x) ((x)& -(x))
using namespace std;
typedef long long ll;
const int N=1e6+10;
const int mod=1e9+7;
int tr[N]={0ll};

int ksm(int m,int k,int mod)
{
	int ans=1;
	int t=m;
	while(k)
	{
		if(k&1) ans=ans*t%mod;
		t=t*t%mod;
		k>>=1;
	}
	return ans;
}



void solve()
{
	ldou n;
	cin>>n;
	ldou a=1.0;
	ldou b=1.0;
	for(ldou i=1.0;i<=n;i+=1.0){
		a*=i;
		b=b*n;
	}
	printf("%.12Lf",a/b*a);
}

signed main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	int tc=1;
	//init(5e5+10);
	//cin>>tc;
	while(tc--)
	{
		solve();
	}
	return 0;
}


int gcd(int a,int b)
{
	return b==0?a:gcd(b,a%b);
}

int exgcd(int a,int b,__int128& x,__int128& y)
{
	if(b==0){
		x=1,y=0;
		return a;//返回值为gcd（a，b）
	}
	int res=exgcd(b,a%b,x,y);//迭代到最底层
	int t=x;
	x=y;
	y=t-(a/b)*y;
	//完成实现时的x1=y2,y1=x2-(a/b)*y2
	//由于是从最深层实现的，故直接用公式向上层传递即可
	return res;
}