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: Wrong Answer)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <ctype.h>
#include <math.h>
#define MAX(X, Y) (((X) > (Y))? (X) : (Y))
#define MIN(X, Y) (((X) < (Y))? (X) : (Y))
int compare_int(const void *a, const void *b){
  return *(long int*)b - *(long int*)a; // 大きい順
}
int compare_int_small(const void *a, const void *b){
  return *(long int*)a - *(long int*)b; // 小さい順
}
// 公約数
long long int gcd(long long int a, long long int b){
  if(b == 0){
    return a;
  }else{
    return gcd(b, a % b);
  }
}
// 最小公倍数 = a * b / 最大公約数
// 公倍数
long long int lcm(long long int a, long long int b){
  return a * b / gcd(a, b);
}

#define TRUE 1
#define FALSE 0
#define MAX_NUM 100002
typedef struct pp{
  char c;
  int h;
  int w;
}P;

int top = 0;
P *Stack = NULL;
P *tmp = NULL;


void initialize(){
  Stack = (P *)calloc(MAX_NUM, sizeof(P));
  top = 0;
}

void finalize(){
  if(Stack != NULL){
    free(Stack);
    Stack = NULL;
  }
}

int isEmpty(){
  if(top <= 0){
    return TRUE;
  }else{
    return FALSE;
  }
}

int isFull(){
  if(top > MAX_NUM - 1){
    return TRUE;
  }else{
    return FALSE;
  }
}

int push(P x){
  fprintf(stderr, "pushするのは%c (h, w) = (%d, %d)\n", x.c, x.h, x.w);
  //printf("push x = %d\n",x);
  if(isFull() != FALSE){
    //    fprintf(stderr, "pushしたのは%d top = %d\n", Stack[top], top);
    printf("Error! Overfllow\n");
    return 0;
  }else{
    top++;
    if(Stack != NULL){
      Stack[top] = x;
    }
    return 0;
  }
}

P pop(){
  if(isEmpty() != FALSE){
    printf("Error! Underfllow\n");
    return *tmp;
  }else{
    top--;
    // return では取りだす値を返す
    //    fprintf(stderr, "popしたのは%d top = %d\n", Stack[top + 1], top);
    return Stack[top+1];
  }
}

long double distance = 0;//そこまでの距離
int n = 0;
int x[10];
int y[10];
int dx = 0, dy = 0;
int now = 0; //現在地
int check_town[10]; //訪れたかどうか
int root[10]; //どの順番か


void check(int *check_town, int *root, int now){
  //  fprintf(stderr, "now = %d enter check()\n", now);
  if(now == n){
    for(int i = 1; i < n; i++)
      {
	fprintf(stderr, "%d",  root[i]);
	dx = x[root[i]] - x[root[i-1]];
	dy = y[root[i]] - y[root[i-1]];
      	distance += sqrt((dx * dx) + (dy * dy));
      }
    fprintf(stderr, "\n");

    fprintf(stderr, "now = %d distance = %Lf\n", now, distance);
    return;
  }else{
    
    // for文で少しずつ考えていく
    for(int i = 0; i < n; i++){//値を一つずつ比較する時に使用
      for(int j = 0; j <= now; j++){ //すでに配列に格納されている所まで
	if(root[j] !=  i){ //使用したことない値ならば
	  if(j == now){
	    root[now] = i;
	    now++;
	    check(check_town, root, now);
	    now--;
	  }
	}else{
	  break;
	}
      }
    }
  }
}




// min
int main(void){

  double d[30500];
  long double ans = 0;
  int pattern = 1;


  scanf("%d", &n);

  for(int i = 0; i < n; i++){
    scanf("%d%d", &x[i], &y[i]);
  }

  // 後で通ったか確認する時に0だと比較できないので、ありえない値を代入しておく
  for(int i = 0; i < n; i++){
    root[i] = 11;
  }
  
  check(check_town, root, now);  
  
  // 経路の数
  for(int i = 1; i <= n; i++)
  {
  	pattern *= i;
  }
  fprintf(stderr, "distance = %Lf pattern = %d\n", distance, pattern);
  // 経路の数で割る
  ans = distance / pattern;
  
  printf("%Lf\n", ans);

  return 0;

}