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)

fn main() {
    let s = std::io::stdin();
    let mut sc = Scanner { stdin: s.lock() };

    let n: usize = sc.read();
    let m: usize = sc.read();
    let mut graph = vec![vec![]; n];
    let mut inverse = vec![vec![]; n];
    for _ in 0..m {
        let a = sc.read::<usize>() - 1;
        let b = sc.read::<usize>() - 1;
        graph[a].push(b);
        inverse[b].push(a);
    }

    let probability = calc_probability(&graph, &inverse, (n, n));
    let mut ans = probability[0];
    for from in 0..n {
        if graph[from].len() == 1 {
            continue;
        }
        if let Some(&next) = graph[from]
            .iter()
            .max_by(|&&next1, &&next2| probability[next1].partial_cmp(&probability[next2]).unwrap())
        {
            let p = calc_probability(&graph, &inverse, (from, next));
            if p[0] < ans {
                ans = p[0];
            }
        }
    }

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

fn calc_probability(
    graph: &Vec<Vec<usize>>,
    inverse: &Vec<Vec<usize>>,
    prohibited: (usize, usize),
) -> Vec<f64> {
    let n = inverse.len();
    let mut dp = vec![0.0; n];
    for next in (0..n).rev() {
        for &from in inverse[next].iter() {
            if (from, next) == prohibited {
                continue;
            }
            let s = if from == prohibited.0 {
                graph[from].len() as f64 - 1.0
            } else {
                graph[from].len() as f64
            };
            dp[from] += (dp[next] + 1.0) / s;
        }
    }
    //    println!("{:?} dp={:?}", prohibited, dp);
    dp
}

pub struct Scanner<R> {
    stdin: R,
}

impl<R: std::io::Read> Scanner<R> {
    pub fn read<T: std::str::FromStr>(&mut self) -> T {
        use std::io::Read;
        let buf = self
            .stdin
            .by_ref()
            .bytes()
            .map(|b| b.unwrap())
            .skip_while(|&b| b == b' ' || b == b'\n')
            .take_while(|&b| b != b' ' && b != b'\n')
            .collect::<Vec<_>>();
        unsafe { std::str::from_utf8_unchecked(&buf) }
            .parse()
            .ok()
            .expect("Parse error.")
    }
    pub fn vec<T: std::str::FromStr>(&mut self, n: usize) -> Vec<T> {
        (0..n).map(|_| self.read()).collect()
    }
    pub fn chars(&mut self) -> Vec<char> {
        self.read::<String>().chars().collect()
    }
}