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 (Clang 3.8.0) (judge verdict: Accepted)

//██████╗ ██╗   ██╗██╗      ██████╗ ███████╗
//██╔══██╗██║   ██║██║      ██╔══██╗██╔════╝
//██████╔╝██║   ██║██║█████╗██║  ██║█████╗
//██╔══██╗██║   ██║██║╚════╝██║  ██║██╔══╝
//██║  ██║╚██████╔╝██║      ██████╔╝███████╗
//╚═╝  ╚═╝ ╚═════╝ ╚═╝      ╚═════╝ ╚══════╝
#include <iostream>
#include <random>
#include <vector>
#include <set>
#include <algorithm>
#include <iomanip>
#include <chrono>
#include <cstdio>


// #pragma GCC target("sse,sse2,sse3,ssse3,sse4")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;

#define fst first
#define snd second
#define mp make_pair
#define mt make_tuple
#define pb push_back
#define eb emplace_back

#define all(v) (v).begin(), (v).end()
#define sz(v) ((int)(v).size())
#define sqr(x) ((x) * (x))

typedef long long ll;
typedef long double ld;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef vector<int> vi;

#define next ajksdslk
#define prev aklsfjk

mt19937_64 mt_rand(chrono::system_clock::now().time_since_epoch().count());

template<typename T1, typename T2> inline bool upmax(T1& a, T2 b) { return (a < b ? (a = b, true) : false); }
template<typename T1, typename T2> inline bool upmin(T1& a, T2 b) { return (b < a ? (a = b, true) : false); }

const int maxn = (int) 1e6 + 100;
const int maxlog = 21;
const int base = 1e9 + 7;
const ld eps = (ld) 1e-9;
const ld PI = acos(-1.);
//const int pp = 41;
const int inf = 2e9;
const ll llinf = 1e18;

ld dist(pii a, pii b) {
  return sqrt( sqr(a.first - b.first) + sqr(a.second - b.second));
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
#ifdef LOCAL
  freopen ("input.txt", "r", stdin);
#endif

  int n; cin >> n;
  vector<pii> a(n);
  for (int i =0 ; i < n; i ++) {
    cin >> a[i].first >>  a[i].second;
  }
  vector<int> p(n);
  iota(all(p), 0);
  ld ans = 0;
  int cnt = 0;
  do {

    for (int i = 0; i + 1 < n; i ++) {
      ans += dist( a[p[i]], a[p[i + 1]] );
    }
    cnt ++;

  } while(next_permutation(all(p)));
  cout << fixed << setprecision(10) << ans / cnt << '\n';
  return 0;
}