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)

use std::io::Read;

fn run() {
    let mut s = String::new();
    std::io::stdin().read_to_string(&mut s).unwrap();
    let mut it = s.trim().split_whitespace();
    let n: usize = it.next().unwrap().parse().unwrap();
    let m: usize = it.next().unwrap().parse().unwrap();
    let mut g = vec![vec![]; n];
    for _ in 0..m {
        let s: usize = it.next().unwrap().parse::<usize>().unwrap() - 1;
        let t: usize = it.next().unwrap().parse::<usize>().unwrap() - 1;
        g[s].push(t);
    }
    let mut dp = vec![0.0; n];
    for v in (0..(n - 1)).rev() {
        dp[v] = g[v].iter().fold(0.0, |s, u| s + dp[*u]) / g[v].len() as f64 + 1.0;
    }
    let mut prob = vec![0.0; n];
    prob[0] = 1.0;
    for v in 0..(n - 1) {
        let p = prob[v] / (g[v].len() as f64);
        for &u in &g[v] {
            prob[u] += p;
        }
    }
    let mut ans = dp[0];
    for v in 0..(n - 1) {
        if g[v].len() == 1 {
            continue;
        }
        let d = g[v].len() as f64;
        let s = (dp[v] - 1.0) * d;
        let p = 1.0 / (d - 1.0);
        let a = dp[0] - prob[v] * dp[v];
        for &u in &g[v] {
            let t = a + prob[v] * (1.0 + (s - dp[u]) * p);
            ans = ans.min(t);
        }
    }
    println!("{:.9}", ans);
}

fn main() {
    run();
}