Problem:
Score : 300 points

Problem Statement

There are N towns in a coordinate plane. Town i is located at coordinates (x_i, y_i). The distance between Town i and Town j is \sqrt{\left(x_i-x_j\right)^2+\left(y_i-y_j\right)^2}.

There are N! possible paths to visit all of these towns once. Let the length of a path be the distance covered when we start at the first town in the path, visit the second, third, \dots, towns, and arrive at the last town (assume that we travel in a straight line from a town to another). Compute the average length of these N! paths.

Constraints

2 \leq N \leq 8

-1000 \leq x_i \leq 1000

-1000 \leq y_i \leq 1000

\left(x_i, y_i\right) \neq \left(x_j, y_j\right) (if i \neq j)

(Added 21:12 JST) All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
x_1 y_1
:
x_N y_N

Output

Print the average length of the paths.
Your output will be judges as correct when the absolute difference from the judge's output is at most 10^{-6}.

Sample Input 1

3
0 0
1 0
0 1

Sample Output 1

2.2761423749

There are six paths to visit the towns: 1 → 2 → 3, 1 → 3 → 2, 2 → 1 → 3, 2 → 3 → 1, 3 → 1 → 2, and 3 → 2 → 1.

The length of the path 1 → 2 → 3 is \sqrt{\left(0-1\right)^2+\left(0-0\right)^2} + \sqrt{\left(1-0\right)^2+\left(0-1\right)^2} = 1+\sqrt{2}.

By calculating the lengths of the other paths in this way, we see that the average length of all routes is:

\frac{\left(1+\sqrt{2}\right)+\left(1+\sqrt{2}\right)+\left(2\right)+\left(1+\sqrt{2}\right)+\left(2\right)+\left(1+\sqrt{2}\right)}{6} = 2.276142...

Sample Input 2

2
-879 981
-866 890

Sample Output 2

91.9238815543

There are two paths to visit the towns: 1 → 2 and 2 → 1. These paths have the same length.

Sample Input 3

8
-406 10
512 859
494 362
-955 -475
128 553
-986 -885
763 77
449 310

Sample Output 3

7641.9817824387

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

/*by freesteed*/
#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define pb push_back
#define fill(x,c) memset(x,c,sizeof(x))
#define fi first
#define se second
#define sz(A) (int)(A.size())
#define all(x) (x).begin(),(x).end()
#define caset int ___T; cin>>___T; for(int cs=1;cs<=___T;cs++)
#define get_be(c,x) (lower_bound(c.begin(),c.end(),x)-c.begin())
//get_be biger & equal
typedef long long ll;
typedef pair<int,int> PII;
typedef vector<int> VI;
const ll mod=1e9+7;
const ll oo=1e9+10;
const int Max = 1e6+10;
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll n;
double tol,ans;
void dfs(vector<pair<double,double> > &p,VI &vis,int s,int cnt,double sum){
	if(cnt==n){
		tol+=sum;
		ans++;
	}
	for(int i=0;i<n;i++){
		if(!vis[i]){
			vis[i] = 1;
			double dis = sqrt((p[s].fi-p[i].fi)*(p[s].fi-p[i].fi)+(p[s].se-p[i].se)*(p[s].se-p[i].se));
			dfs(p,vis,i,cnt+1,sum+dis);
			vis[i] = 0;
		}
	}
}
void solve(){
	cin>>n;
	vector<pair<double,double> > p;
	rep(i,0,n){
		double x,y;
		cin>>x>>y;
		p.pb({x,y});
	}
	rep(i,0,n)
	{
		VI vis(n,0);
		vis[i] = 1;
		dfs(p,vis,i,1,0.0);
	}
	printf("%.9f",tol/ans);
}
int main() {
	//ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	//caset{
		solve();
	//}
  	return 0;
}