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)

#![allow(unused_imports, unused_macros, dead_code)]

macro_rules! min {
    (.. $x:expr) => {{
        let mut it = $x.iter();
        it.next().map(|z| it.fold(z, |x, y| min!(x, y)))
    }};
    ($x:expr) => ($x);
    ($x:expr, $($ys:expr),*) => {{
        let t = min!($($ys),*);
        if $x < t { $x } else { t }
    }}
}
macro_rules! max {
    (.. $x:expr) => {{
        let mut it = $x.iter();
        it.next().map(|z| it.fold(z, |x, y| max!(x, y)))
    }};
    ($x:expr) => ($x);
    ($x:expr, $($ys:expr),*) => {{
        let t = max!($($ys),*);
        if $x > t { $x } else { t }
    }}
}


macro_rules! ewriteln {
    ($($args:expr),*) => { let _ = writeln!(&mut std::io::stderr(), $($args),*); };
}
macro_rules! trace {
    ($x:expr) => { ewriteln!(">>> {} = {:?}", stringify!($x), $x) };
    ($($xs:expr),*) => { trace!(($($xs),*)) }
}
macro_rules! put {
    (.. $x:expr) => {{
        let mut it = $x.iter();
        if let Some(x) = it.next() { print!("{}", x); }
        for x in it { print!(" {}", x); }
        println!("");
    }};
    ($x:expr) => { println!("{}", $x) };
    ($x:expr, $($xs:expr),*) => { print!("{} ", $x); put!($($xs),*) }
}

const M: i64 = 1_000_000_007;

fn main() {
    let mut sc = Scanner::new();
    let n: usize = sc.cin();
    let m: usize = sc.cin();
    let mut g = vec![vec![]; n];
    for _ in 0..m {
        let s = sc.cin::<usize>() - 1;
        let t = sc.cin::<usize>() - 1;
        g[s].push(t);
    }
    let mut ans;
    {
        let mut dp = vec![0.0f64; n];
        for i in (0..n-1).rev() {
            let deg = g[i].len();
            let p = 1.0 / (deg as f64);
            dp[i] = 1.0;
            for &j in g[i].iter() {
                dp[i] += dp[j] * p;
            }
        }
        ans = dp[0];
    }
    for sx in 0..n {
        let mut dp = vec![0.0f64; n];
        for i in (0..n-1).rev() {
            if i == sx && g[i].len() > 1 {
                let v = g[i].iter().map(|&j| (dp[j], j)).collect::<Vec<_>>();
                let &(_, tx) = max!(..v).unwrap();
                let deg = g[i].len() - 1;
                let p = 1.0 / (deg as f64);
                dp[i] = 1.0;
                for &j in g[i].iter() {
                    if j == tx { continue }
                    dp[i] += dp[j] * p;
                }
            } else {
                let deg = g[i].len();
                let p = 1.0 / (deg as f64);
                dp[i] = 1.0;
                for &j in g[i].iter() {
                    dp[i] += dp[j] * p;
                }
            }
        }
        ans = min!(ans, dp[0]);
    }
    put!(ans);
}

use std::io::{self, Write};
use std::str::FromStr;
use std::collections::VecDeque;

struct Scanner { stdin: io::Stdin, buffer: VecDeque<String>, }
impl Scanner {
    fn new() -> Self { Scanner { stdin: io::stdin(), buffer: VecDeque::new() } }
    fn cin<T: FromStr>(&mut self) -> T {
        while self.buffer.len() == 0 {
            let mut line = String::new();
            let _ = self.stdin.read_line(&mut line);
            for w in line.split_whitespace() {
                self.buffer.push_back(String::from(w));
            }
        }
        self.buffer.pop_front().unwrap().parse::<T>().ok().unwrap()
    }
    fn chars(&mut self) -> Vec<char> { self.cin::<String>().chars().collect() }
    fn vec<T: FromStr>(&mut self, n: usize) -> Vec<T> { (0..n).map(|_| self.cin()).collect() }
}