Problem:
Score : 600 points

Problem Statement

There is a cave consisting of N rooms and M one-directional passages. The rooms are numbered 1 through N.

Takahashi is now in Room 1, and Room N has the exit. The i-th passage connects Room s_i and Room t_i (s_i < t_i) and can only be traversed in the direction from Room s_i to Room t_i. It is known that, for each room except Room N, there is at least one passage going from that room.

Takahashi will escape from the cave. Each time he reaches a room (assume that he has reached Room 1 at the beginning), he will choose a passage uniformly at random from the ones going from that room and take that passage.

Aoki, a friend of Takahashi's, can block one of the passages (or do nothing) before Takahashi leaves Room 1. However, it is not allowed to block a passage so that Takahashi is potentially unable to reach Room N.

Let E be the expected number of passages Takahashi takes before he reaches Room N. Find the value of E when Aoki makes a choice that minimizes E.

Constraints

2 \leq N \leq 600

N-1 \leq M \leq \frac{N(N-1)}{2}

s_i < t_i

If i != j, (s_i, t_i) \neq (s_j, t_j). (Added 21:23 JST)

For every v = 1, 2, ..., N-1, there exists i such that v = s_i.

Input

Input is given from Standard Input in the following format:

N M
s_1 t_1
:
s_M t_M

Output

Print the value of E when Aoki makes a choice that minimizes E.
Your output will be judged as correct when the absolute or relative error from the judge's output is at most 10^{-6}.

Sample Input 1

4 6
1 4
2 3
1 3
1 2
3 4
2 4

Sample Output 1

1.5000000000

If Aoki blocks the passage from Room 1 to Room 2, Takahashi will go along the path 1 → 3 → 4 with probability \frac{1}{2} and 1 → 4 with probability \frac{1}{2}. E = 1.5 here, and this is the minimum possible value of E.

Sample Input 2

3 2
1 2
2 3

Sample Output 2

2.0000000000

Blocking any one passage makes Takahashi unable to reach Room N, so Aoki cannot block a passage.

Sample Input 3

10 33
3 7
5 10
8 9
1 10
4 6
2 5
1 7
6 10
1 4
1 3
8 10
1 5
2 6
6 9
5 6
5 8
3 6
4 8
2 7
2 9
6 7
1 2
5 9
6 8
9 10
3 9
7 8
4 5
2 10
5 7
3 5
4 7
4 9

Sample Output 3

3.0133333333

Solution in Rust (1.15.1) (judge verdict: Wrong Answer)

use std::cmp::{Ordering, min};
use std::io::{self, BufRead};
use std::ops::{Add, Mul};

fn gcd(x: usize, y: usize) -> usize {
  let (mut a, mut b) = if x < y {
    (y, x)
  } else {
    (x, y)
  };
  while b != 0 {
    let r = a % b;
    a = b;
    b = r;
  }
  a
}

fn lcm_by_gcd(x: usize, y: usize, g: usize) -> usize {
  x * y / g
}

#[derive(Debug, Clone, Copy, Eq)]
struct Rational(usize, usize);

impl Add for Rational {
  type Output = Self;
  fn add(self, rhs: Self) -> Self::Output {
    let g = gcd(self.1, rhs.1);
    let l = lcm_by_gcd(self.1, rhs.1, g);
    Rational(self.0 * rhs.1 / g + rhs.0 * self.1 / g, l)
  }
}

impl Mul for Rational {
  type Output = Self;
  fn mul(self, rhs: Self) -> Self::Output {
    let n = self.0 * rhs.0;
    let d = self.1 * rhs.1;
    let g = gcd(n, d);
    Rational(n / g, d / g)
  }
}

impl Ord for Rational {
  fn cmp(&self, rhs: &Self) -> Ordering {
    let g = gcd(self.1, rhs.1);
    let l = self.0 * rhs.1 / g;
    let r = rhs.0 * self.1 / g;
    l.cmp(&r)
  }
}

impl PartialOrd for Rational {
  fn partial_cmp(&self, rhs: &Self) -> Option<Ordering> {
    Some(self.cmp(rhs))
  }
}

impl PartialEq for Rational {
  fn eq(&self, rhs: &Self) -> bool {
    let g = gcd(self.1, rhs.1);
    let l = self.0 * rhs.1 / g;
    let r = rhs.0 * self.1 / g;
    l == r
  }
}

const ZERO: Rational = Rational(0, 1);
const ONE: Rational = Rational(1, 1);

fn main() {
  let stdin = io::stdin();
  let mut lines = stdin.lock().lines();
  let line = lines.next().unwrap().unwrap();
  let vs: Vec<usize> = line.split(' ').map(|x| x.parse().unwrap()).collect();
  let (n, m) = (vs[0], vs[1]);
  let mut es: Vec<Vec<usize>> = vec![Vec::new(); n];
  for _ in 0..m {
    let line = lines.next().unwrap().unwrap();
    let vs: Vec<usize> = line.split(' ').map(|x| x.parse().unwrap()).collect();
    let (s, t) = (vs[0] - 1, vs[1] - 1);
    es[s].push(t);
  }
  for i in 0..n {
    es[i].sort();
  }
  let mut memo0: Vec<Rational> = vec![ZERO; n];
  let mut memo1: Vec<Option<Rational>> = vec![None; n];
  for k in 1..n {
    let i = n - k - 1;
    let count = es[i].len();
    memo0[i] = ONE + Rational(1, count) * es[i].iter().map(|j| memo0[*j]).fold(ZERO, |acc, x| acc + x);
    if count > 1 {
      let mut temp = Vec::new();
      for j in 0..count {
        let x = (0..count).filter(|l| *l != j).map(|l| memo0[es[i][l]]).fold(ZERO, |acc, x| acc + x);
        let r1 = ONE + Rational(1, count - 1) * x;
        temp.push(r1);
        if let Some(y) = memo1[es[i][j]] {
          let r2 = ONE + Rational(1, count) * (x + y);
          temp.push(r2);
        }
      }
      memo1[i] = Some(*temp.iter().min().unwrap());
    }
  }
  let r = if let Some(r) = memo1[0] {
    min(memo0[0], r)
  } else {
    memo0[0]
  };
  let ans = r.0 as f64 / r.1 as f64;
  println!("{}", ans);
}