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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <string.h>
#include <stdbool.h>
#include <stdint.h>
#include <stdlib.h>
#include <limits.h>
#include <math.h>
#include <assert.h>
typedef int64_t ll;
typedef uint64_t ull;
int acs(const void *a, const void *b){return *(int*)a - *(int*)b;} /* 1,2,3,4.. */
int des(const void *a, const void *b){return *(int*)b - *(int*)a;} /* 8,7,6,5.. */
#define min(a,b) (a < b ? a: b)
#define max(a,b) (a > b ? a: b)

#define MAXN (8)
#define MOD (1000000007)

// べき乗余
int64_t modpow(int64_t b, int64_t e)
{
    int64_t result = 1;
    while(e>0)
    {
        if(e&1) result = (result * b) % MOD;
        e >>= 1;
        b = (b * b) % MOD;
    }
    return result;
}


// 組み合わせ数
// nCk % MOD
ll cf[MAXN+1]; // 1*2*3*...*i
ll ci[MAXN+1]; // 1/ 1*2*3*...*i
void CombInit(ll n)
{
    cf[0] = 1;
    for(int i=1;i<=n;i++) cf[i] = cf[i-1] * i %MOD;
    ci[n] = modpow(cf[n],MOD-2);
    for(int i=n-1;i>=0;i--) ci[i] = ci[i+1] * (i+1) %MOD;
}
ll Comb(ll a,ll b)
{
    return cf[a]*ci[b]%MOD*ci[a-b]%MOD;
}
// カタラン数 2nまで参照することに注意
ll Catalan(ll n)
{
    return cf[2*n]*ci[n+1]%MOD*ci[n]%MOD;
}

int x[MAXN];
int y[MAXN];

int main(void)
{
    ll n;
    scanf("%ld",&n);
    CombInit(n);
    for(int i=0;i<n;i++) scanf("%d %d",&x[i],&y[i]);


    double ans = 0;
    for(int i=0;i<n;i++) for(int j=i+1;j<n;j++)
    {
        double dis;
        dis = sqrt((double)((x[i]-x[j])*(x[i]-x[j]) +(y[i]-y[j])*(y[i]-y[j])));
        ans += dis*cf[n-1]*2;
    }
    printf("%.10lf\n", ans/cf[n]);
    return 0;
}