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: Accepted)

#![allow(unused_imports)]
#![allow(non_snake_case)]
use std::cmp::*;
use std::collections::*;
use std::io::Write;

#[allow(unused_macros)]
macro_rules! debug {
    ($($e:expr),*) => {
        #[cfg(debug_assertions)]
        $({
            let (e, mut err) = (stringify!($e), std::io::stderr());
            writeln!(err, "{} = {:?}", e, $e).unwrap()
        })*
    };
}

fn main() {
    let v = read_vec::<usize>();
    let (n, m) = (v[0], v[1]);
    let mut edges = vec![Vec::new(); n];
    for i in 0..m {
        let v = read_vec::<usize>();
        let (s, t) = (v[0] - 1, v[1] - 1);
        edges[s].push(t);
    }

    let solve = |diff: usize| {
        let mut dp = vec![0.0; n];
        for i in (0..n - 1).rev() {
            let coef = 1.0 / edges[i].len() as f64;
            for &to in edges[i].iter() {
                dp[i] += (1.0 + dp[to]) * coef;
            }
            if i == diff {
                dp[i] += 1.0;
            }
        }
        dp
    };
    let dp = solve(n);
    // debug!(dp);
    let mut ans = dp[0];
    for i in 0..n - 1 {
        if edges[i].len() <= 1 {
            continue;
        }
        let coef = if i == 0 { 1.0 } else { solve(i)[0] - dp[0] };
        let len = edges[i].len() as f64;
        for &to in edges[i].iter() {
            let temp1 = dp[i] - (1.0 + dp[to]) / edges[i].len() as f64;
            let temp2 = temp1 * len / (len - 1.0);
            let diff = temp2 - dp[i];
            // debug!(i, coef, to, diff, temp1, temp2, dp[0] + diff * coef);
            ans = min(Total(ans), Total(dp[0] + diff * coef)).0;
        }
    }
    println!("{}", ans);
}

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()
}

#[derive(PartialEq, PartialOrd)]
pub struct Total<T>(pub T);

impl<T: PartialEq> Eq for Total<T> {}

impl<T: PartialOrd> Ord for Total<T> {
    fn cmp(&self, other: &Total<T>) -> std::cmp::Ordering {
        self.0.partial_cmp(&other.0).unwrap()
    }
}