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 MOD 1000000007
#define pi 3.1415926535897932384626433832795028841971693993751058209749445923078164062
#define ll long long
#define pb push_back
#define pf push_front
#define mp make_pair
#define mt make_tuple
#define ub upper_bound
#define lb lower_bound
#define popb pop_back()
#define popf pop_front()
#define ff first
#define ss second
#define vl vector<ll>
#define vi vector<int>
#define vs vector<string>
#define vll vector< pair<ll,ll> >
#define vii vector< pair<int,int> >
#define viii vector< tuple <int,int,int> >
#define vlll vector< tuple <ll,ll,ll> >
#define vvl vector<vector<ll>>
#define vv vector<vector<int>>
#define all(v) v.begin(),v.end()
#define sqrt sqrtl
#define cbrt cbrtl
#define pll pair<ll,ll>
#define pii pair<int,int>
#define mapcl map<char,ll>
#define mapci map<char,int>
#define mapll map<ll,ll>
#define mapii map<int,int>
#define seti set<int>
ifstream fin("input.txt");
ofstream fout("output.txt");
#define FOR(i, l, r) for (int i = int(l); i < int(r); ++i)

ll fast_exp(ll base, ll exp) {ll res=1;while(exp>0) {if(exp%2==1) res=(res*base)%MOD;base=(base*base)%MOD;exp/=2;}return res%MOD;}
int gcd(int a,int b){while (a&&b)a>b?a%=b:b%=a;return a+b;}
int val(char c){if (c >= '0' && c <= '9')return (int)c - '0';else return (int)c - 'A' + 10;}
ll pows(int a , int b){ll res=1;for(int i=0; i<b; ++i){res*=a;}return res;}
ll logx(ll base, ll num){int cnt=0;while(num!=1){num/=base; ++cnt;}return cnt;}
ll divisibles(ll a, ll b, ll m){if(a%m==0)return (b/m)-(a/m)+1;else return (b/m)-(a/m);}// in [a,b]
string bitstring(int n){string s;while(n){s+=(n%2)+'0';n/=2;}reverse(all(s));return s;} // only sufficient bits
// dsu start

vi root(200001,0);
vi size(200001,1);
int find(int x){while(x!=root[x])x = root[x];return x;}
bool same(int a,int b){return find(a)==find(b);}
void unite(int a, int b){a = find(a);b = find(b);if(size[a]<size[b])swap(a,b);size[a] += size[b];root[b] = a;}

// dsu end
vi vis(200001,0);
vi adj[200001];
set< pair<double,double> > setx;
vector < pair <double,double> >  vec;
vector <double> ans;
int n;
mapii mps;
void pathgen(int s, int len, double sum)
{
	if(len==n-1)
	{
		//cout<<"YAY";
		ans.pb(sum);
		return ;
	}
	for(int i=0; i<n; ++i)
	{
		if(!mps[i])
		{
			++mps[i];
			//cout<<s<<" "<<i<<"\n";
			double x = vec[s].ff-vec[i].ff;
			double y = vec[s].ss-vec[i].ss;
			double ans = sqrtl((pow(x,2)) + pow(y,2));
			//cout<<ans<<"\n";
			pathgen(i,len+1,sum+ans);
			--mps[i];
		}
	}


}
int main()
{	
    std::ios::sync_with_stdio(false);
    //string bitstring = std::bitset< 3 >( 7 ).to_string(); <bits> (num)
    //srand(time(0));
    //cin.tie(NULL);
    //cout.tie(NULL);
 	
 	cin>>n;
 	for(int i=0; i<n; ++i)
 	{
 		double x,y;
 		cin>>x>>y;
 		vec.pb(mp(x,y));
 	}
 	for(int i=0; i<n; ++i)
 	{
 		++mps[i];
 		pathgen(i,0,0);
 		--mps[i];
 	}
 	double sum = 0;
 	for(auto x : ans)
 		sum+=x;
 	//cout<<fixed<<sum<<"\n";
 	sum/=ans.size();
 	cout<<fixed<<sum<<"\n";

}