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: Runtime Error)

// This code is generated by [cargo-atcoder](https://github.com/tanakh/cargo-atcoder)
// Original source code:
/*
use proconio::input;
#[allow(unused_imports)]
use proconio::marker::*;

fn main() {
    input! {
        n: usize, m: usize,
        mut e: [(Usize1, Usize1); m]
    }
    e.sort();
    let mut tree = vec![vec![]; n];
    for &(s, t) in &e {
        tree[s].push(t);
    }

    fn dfs(memo: &mut [(f64, Vec<f64>)], tree: &[Vec<usize>], i: usize) {
        if i == memo.len() - 1 {
            return;
        }
        let mut s = 0.;
        let mut v = vec![];
        for &t in &tree[i] {
            if memo[t].0 < 0. {
                dfs(memo, tree, t);
            }
            let x = memo[t].0 + 1.;
            v.push(x);
            s += x;
        }
        let res = s as f64 / tree[i].len() as f64;
        memo[i].0 = res;
        memo[i].1 = v;
    }
    let mut memo = vec![(-1., vec![]); n];
    memo[n - 1].0 = 0.;
    dfs(&mut memo, &tree, 0);

    let mut v_rate = vec![0.; n];
    v_rate[0] = 1.0f64;
    let mut res = memo[0].0;
    for i in 0..n {
        if tree[i].len() == 1 {
            v_rate[tree[i][0]] += v_rate[i];
            continue;
        }
        for j in 0..tree[i].len() {
            v_rate[tree[i][j]] += v_rate[i] / tree[i].len() as f64;
            let x = (memo[i].0 * (tree[i].len() as f64) - (memo[i].1)[j])
                / ((tree[i].len() - 1) as f64);
            res = res.min(memo[0].0 - (memo[i].0 - x) * v_rate[i]);
            // println!("{} {} {} {}", i, j, x, memo[0].0 - (memo[i].1)[j] * r);
        }
    }
    // println!("{:?}", memo);
    println!("{}", res);
}
*/

fn main() {
    let exe = "/tmp/binF70F397F";
    std::io::Write::write_all(&mut std::fs::File::create(exe).unwrap(), &decode(BIN)).unwrap();
    std::fs::set_permissions(exe, std::os::unix::fs::PermissionsExt::from_mode(0o755)).unwrap();
    std::process::exit(std::process::Command::new(exe).status().unwrap().code().unwrap())
}

fn decode(v: &str) -> Vec<u8> {
    let mut ret = vec![];
    let mut buf = 0;
    let mut tbl = vec![64; 256];
    for i in 0..64 { tbl[TBL[i] as usize] = i as u8; }
    for (i, c) in v.bytes().filter_map(|c| { let c = tbl[c as usize]; if c < 64 { Some(c) } else { None } }).enumerate() {
        match i % 4 {
            0 => buf = c << 2,
            1 => { ret.push(buf | c >> 4); buf = c << 4; }
            2 => { ret.push(buf | c >> 2); buf = c << 6; }
            3 => ret.push(buf | c),
            _ => unreachable!(),
        }
    }
    ret
}

const TBL: &'static [u8] = b"ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";
const BIN: &'static str = "<base64-encoded ELF executable removed>";