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;
use std::cmp;

fn main() {
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();
    let mut iter = buf.split_whitespace();

    let n: usize = iter.next().unwrap().parse().unwrap();
    let m: usize = iter.next().unwrap().parse().unwrap();

    let mut g: Vec<Vec<usize>> = vec![Vec::new(); n];
    let mut s: Vec<usize> = Vec::with_capacity(m);
    let mut t: Vec<usize> = Vec::with_capacity(m);
    for i in 0..m {
        s.push(iter.next().unwrap().parse().unwrap());
        t.push(iter.next().unwrap().parse().unwrap());
        s[i] -= 1;
        t[i] -= 1;
        g[s[i]].push(t[i]);
    }

    let mut arrive: Vec<f32> = vec![0.0; n];
    arrive[0] = 1.0;
    for u in 0..n {
        for v in &g[u] {
            arrive[*v] += arrive[u] / g[u].len() as f32;
        }
    }

    let mut aisle: Vec<f32> = vec![0.0; n];
    for u in (0..n).rev() {
        for v in &g[u] {
            aisle[u] += aisle[*v] + 1.0;
        }
        aisle[u] /= cmp::max(1, g[u].len()) as f32;
    }

    let mut ans: f32 = aisle[0];
    for i in 0..m {
        if g[s[i]].len() == 1 { continue; }

        let mut res: f32 = 0.0;
        res += aisle[s[i]] * g[s[i]].len() as f32;
        res -= aisle[t[i]] + 1.0;
        res /= (g[s[i]].len() - 1) as f32;
        res = aisle[s[i]] - res;
        res = aisle[0] - res * arrive[s[i]];
        ans = f32::min(res, ans);
    }

    println!("{:.10}", ans);
}