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)

#include <bits/stdc++.h>
using namespace std;
#define int long long
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
using P=pair<long long,long long>;
#define rep(i,n) for(long long i=0; i<(long long)n; i++)
#define FOR(i,a,b) for(long long i=a; i<b; i++)
#define ll long long
#define all(x) (x).begin(), (x).end()
#define SZ(x) ((long long)(x).size())
#define COUT(x) cout << x << endl
#define PB(x) push_back(x)
#define MP make_pair
#define vint vector<int>
#define vvint vector<vector<int>>
#define vstr vector<string>
#define vp vector<pair<int,int>>
#define vb vector<bool>
#define SUM(x) accumulate(x.begin(),x.end(),0)
#define MAX(x) *max_element(x.begin(),x.end())
#define MIN(x) *min_element(x.begin(),x.end())
#define couty cout<<'Y'<<'e'<<'s'<<endl
#define coutn cout<<'N'<<'o'<<endl
#define coutY cout<<'Y'<<'E'<<'S'<<endl
#define coutN cout<<'N'<<'O'<<endl
#define yn(x) cout<<(x?"Yes":"No")<<endl
#define YN(x) cout<<(x?"YES":"NO")<<endl
long long gcd(long long a,long long b){return b?gcd(b,a%b):a;}
long long lcm(long long a, long long b){return a*b/gcd(a,b);}
const long long dx[4]={1,0,-1,0};
const long long dy[4]={0,1,0,-1};
const long long INF = 1e12;
const long long MOD = 1e9+7;
#define f first
#define s second
#define ld long double

vint x,y;
ld dist(int i,int j){
    ld dx=x[i]-x[j];
    ld dy=y[i]-y[j];
    return pow(dx*dx+dy*dy,0.5);

}

int fractal(int n){
    int ret=1;
    FOR(i,1,n+1) ret*=i;
    return ret;
}
signed main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout<<fixed<<setprecision(15);

    int n; cin >> n;
    x.resize(n); y.resize(n);
    rep(i,n) cin >> x[i] >> y[i];
    vint order(n);
    rep(i,n) order[i]=i;

    ld res=0.0;

    do{
        rep(i,n-1){
            res+=dist(order[i],order[i+1]);
        }
    }while(next_permutation(all(order)));
    //cout<<fractal(3)<<endl;
    cout<<res/fractal(n)<<endl;
}