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(non_snake_case)]
#![allow(unused_variables)]
#![allow(dead_code)]

fn main() {
    let (N, M): (usize, usize) = {
        let mut line: String = String::new();
        std::io::stdin().read_line(&mut line).unwrap();
        let mut iter = line.split_whitespace();
        (
            iter.next().unwrap().parse().unwrap(),
            iter.next().unwrap().parse().unwrap()
        )
    };
    let (s, t): (Vec<usize>, Vec<usize>) = {
        let (mut s, mut t) = (vec![], vec![]);
        for _ in 0..M {
            let mut line: String = String::new();
            std::io::stdin().read_line(&mut line).unwrap();
            let mut iter = line.split_whitespace();
            s.push(iter.next().unwrap().parse().unwrap());
            t.push(iter.next().unwrap().parse().unwrap());
        }
        (s, t)
    };

    #[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()
        }
    }

    let G = {
        let mut G = vec![std::collections::HashSet::new(); N + 1];
        for i in 0..M { G[s[i]].insert(t[i]); }
        G
    };
    const INF: f64 = 1_000_000_000.0;
    let mut dp = vec![0.0; N + 1];
    for i in (1..N).rev() { dp[i] = G[i].iter().map(|&j| 1.0 + dp[j]).sum::<f64>() / (G[i].len() as f64); }
    let ans = (1..N).map(|k|
        if G[k].len() == 1 { Total(dp[1]) }
        else {
            let l = G[k].iter().map(|&i| i).max_by_key(|&i| Total(dp[i])).unwrap();
            let mut dp2 = vec![0.0; N + 1];
            for i in (1..N).rev() {
                dp2[i] = if i == k {
                    G[i].iter().map(|&j|
                        if j == l { 0.0 }
                        else { 1.0 + dp2[j] }
                    ).sum::<f64>() / ((G[i].len() - 1) as f64)
                } else { G[i].iter().map(|&j| { 1.0 + dp2[j] }).sum::<f64>() / (G[i].len()as f64) };
            }
            Total(dp2[1])
    }).min().unwrap().0;

    println!("{}", ans);
}