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: Time Limit Exceeded)

use std::cmp::{max, min};
use std::collections::HashMap;
use std::io::{stdin, Read};

#[allow(unused_macros)]
macro_rules! parse {
    ($it: ident ) => {};
    ($it: ident, ) => {};
    ($it: ident, $var:ident : $t:tt $($r:tt)*) => {
        let $var = parse_val!($it, $t);
        parse!($it $($r)*);
    };
    ($it: ident, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = parse_val!($it, $t);
        parse!($it $($r)*);
    };
}

#[allow(unused_macros)]
macro_rules! parse_val {
    ($it: ident, [$t:tt; $len:expr]) => {
        (0..$len).map(|_| parse_val!($it, $t)).collect::<Vec<_>>();
    };
    ($it: ident, ($($t: tt),*)) => {
        ($(parse_val!($it, $t)),*)
    };
    ($it: ident, u1) => {
        $it.next().unwrap().parse::<usize>().unwrap() -1
    };
    ($it: ident, $t: ty) => {
        $it.next().unwrap().parse::<$t>().unwrap()
    };
}

fn solve(s: &str) {
    let mut it = s.split_whitespace();
    parse!(it, n: usize, m: usize, mut st: [(u1, u1); m]);
    let mut g = vec![vec![]; n];
    for &(s, t) in st.iter() {
        g[s].push(t)
    }
    let mut dp = vec![0.; n];
    dfs(0, n - 1, &g, &mut dp, (0, 0));
    let mut ret = dp[0];
    let mut dp1 = vec![0.; n];
    for i in 0..(n - 1) {
        let mut j = 0;
        let mut m = 0.;
        for &ij in g[i].iter() {
            if m < dp[ij] {
                j = ij;
                m = dp[j]
            }
        }
        if j != 0 {
            dfs(0, n - 1, &g, &mut dp1, (i, j));
            ret = ret.min(dp1[0]);
        }
    }
    println!("{}", ret);
}
fn dfs(u: usize, dest: usize, g: &Vec<Vec<usize>>, dp: &mut Vec<f64>, stop: (usize, usize)) {
    if u == dest {
        return;
    }
    let mut count = 0.;
    let mut ret = 0.;
    for &v in g[u].iter() {
        if (u, v) == stop {
            continue;
        }
        count += 1.;
        dfs(v, dest, g, dp, stop);
        ret += dp[v];
    }
    dp[u] = ret / count + 1.
}

fn main() {
    let mut s = String::new();
    stdin().read_to_string(&mut s).unwrap();
    solve(&s);
}

#[cfg(test)]
mod tests {
    use super::*;
    #[test]
    fn test_input() {
        let s = "
        ";
        solve(s);
    }
}