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: Wrong Answer)

use std::cmp::Ordering;
use std::io::BufRead;
fn main() {
    let std = std::io::stdin();
    let mut paths: Vec<(usize, usize)> = std
        .lock()
        .lines()
        .map(|l| {
            let st: Vec<usize> = l
                .unwrap()
                .split_whitespace()
                .flat_map(|w| w.parse())
                .collect();
            return (st[0], st[1]);
        })
        .collect();
    let (n, m) = paths.swap_remove(0);
    //println!("{:?},{:?},{:?}", n, m, paths);
    let mut dp: Vec<f64> = vec![0f64; n];
    //i+1->n => 1
    //i+1->x->n = 1+avg
    for i in (0..n - 1).rev() {
        let mut cnt = 0;
        for &(s, e) in &paths {
            if s == i + 1 {
                cnt += 1;
                dp[i] += dp[e - 1];
            }
        }
        dp[i] /= cnt as f64;
        dp[i] += 1f64;
    }
    //println!("{:?}", dp);
    let seal_from = dp
        .iter()
        .map(|d| *d)
        .enumerate()
        .fold((0, 0.0 / 0.0), |l, r| {
            if (l.1).partial_cmp(&(r.1)).unwrap_or(Ordering::Less) == Ordering::Greater {
                l
            } else {
                r
            }
        })
        .0;
    let mut dp_sealed: Vec<f64> = Vec::new();
    dp_sealed.push(dp[0]);
    for seal in paths
        .iter()
        .enumerate()
        .filter(|path| (path.1).0 == seal_from + 1)
    {
        dp = vec![0f64; n];
        for i in (0..n - 1).rev() {
            let mut cnt = 0;
            for &(s, e) in &paths {
                if s == i + 1 && &(s, e) != seal.1 {
                    cnt += 1;
                    dp[i] += dp[e - 1];
                }
            }
            dp[i] /= cnt as f64;
            dp[i] += 1f64;
        }
        dp_sealed.push(dp[0]);
        //println!("sealed:{:?},{:?}", seal, dp);
    }
    //println!("dp_seald{:?}", dp_sealed);
    println!("{}", dp_sealed.into_iter().fold(0.0 / 0.0, f64::min));
}