use std::vec::Vec;

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>().split_whitespace().map(|e| e.parse().ok().unwrap()).collect()
}

fn read_touple<T: std::str::FromStr + Copy>() -> (T, T) {
  let v: Vec<T> = read_vec();
  assert_eq!(v.len(), 2);
  (v[0], v[1])
}

fn main() {
  let n: usize = read();
  let mut xy: Vec<(f64, f64)> = Vec::new();
  for _ in 0..n {
    let t = read_touple::<f64>();
    xy.push(t);
  }
  
  let mut ans: f64 = 0.0;
  let div: f64 = (1..n+1).fold(1.0, |acc, i| acc * i as f64);
  let mut perm: Vec<usize> = (0..n).into_iter().collect();
  ;
  //println!("{}", count);
  //println!("{:?}", xy);
  while  {
    for i in 0..n-1 {
      ans += dist(xy[perm[i]], xy[perm[i+1]]);
    }
    perm.next_permutation()
  }{}
  println!("{}", ans / div as f64);
}

fn dist(x: (f64, f64), y: (f64, f64)) -> f64 {
  ((x.0 - y.0).powi(2) + (x.1 - y.1).powi(2)).sqrt()
}

pub trait LexicalPermutation {
    /// Return `true` if the slice was permuted, `false` if it is already
    /// at the last ordered permutation.
    fn next_permutation(&mut self) -> bool;
    /// Return `true` if the slice was permuted, `false` if it is already
    /// at the first ordered permutation.
    fn prev_permutation(&mut self) -> bool;
}

impl<T> LexicalPermutation for [T] where T: Ord {
    /// Original author in Rust: Thomas Backman <serenity@exscape.org>
    fn next_permutation(&mut self) -> bool {
        // These cases only have 1 permutation each, so we can't do anything.
        if self.len() < 2 { return false; }

        // Step 1: Identify the longest, rightmost weakly decreasing part of the vector
        let mut i = self.len() - 1;
        while i > 0 && self[i-1] >= self[i] {
            i -= 1;
        }

        // If that is the entire vector, this is the last-ordered permutation.
        if i == 0 {
            return false;
        }

        // Step 2: Find the rightmost element larger than the pivot (i-1)
        let mut j = self.len() - 1;
        while j >= i && self[j] <= self[i-1]  {
            j -= 1;
        }

        // Step 3: Swap that element with the pivot
        self.swap(j, i-1);

        // Step 4: Reverse the (previously) weakly decreasing part
        self[i..].reverse();

        true
    }

    fn prev_permutation(&mut self) -> bool {
        // These cases only have 1 permutation each, so we can't do anything.
        if self.len() < 2 { return false; }

        // Step 1: Identify the longest, rightmost weakly increasing part of the vector
        let mut i = self.len() - 1;
        while i > 0 && self[i-1] <= self[i] {
            i -= 1;
        }

        // If that is the entire vector, this is the first-ordered permutation.
        if i == 0 {
            return false;
        }

        // Step 2: Reverse the weakly increasing part
        self[i..].reverse();

        // Step 3: Find the rightmost element equal to or bigger than the pivot (i-1)
        let mut j = self.len() - 1;
        while j >= i && self[j-1] < self[i-1]  {
            j -= 1;
        }

        // Step 4: Swap that element with the pivot
        self.swap(i-1, j);

        true
    }

}

Measured memory_bytes: 4352