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 (1.42.0) (judge verdict: Time Limit Exceeded)

fn main() {
	let mut s = String::new();
	use std::io::Read;
	std::io::stdin().read_to_string(&mut s).unwrap();
	let mut s = s.split_whitespace();
	let n: usize = s.next().unwrap().parse().unwrap();
	let m: usize = s.next().unwrap().parse().unwrap();
	let mut edges = vec![];
	for _ in 0..m {
		let a = s.next().unwrap().parse::<usize>().unwrap() - 1;
		let b = s.next().unwrap().parse::<usize>().unwrap() - 1;
		edges.push((a, b));
	}
	edges.sort();
	let calc = |invalid| {
		let mut dp = vec![(0., 0.); n];
		let mut visit = vec![false; n];
		dp[0].1 = 1.;
		visit[0] = true;

		let mut li = 0;
		let mut ri = 0;
		while li < edges.len() {
			li = ri;
			let src = edges[li].0;
			while ri < edges.len() && src == edges[ri].0 {
				ri += 1;
			}
			let cnt = ri - li - if li <= invalid && invalid < ri { 1 } else { 0 };
			if cnt == 0 {
				dp[n - 1].0 = 1000.;
			}
			let cnt = cnt as f64;
			for i in li..ri {
				if i == invalid {
					continue;
				}
				let to = edges[i].1;
				visit[to] |= visit[src];
				dp[to].0 += (dp[src].0 + dp[src].1) / cnt;
				dp[to].1 += dp[src].1 / cnt;
			}
			li = ri;
		}
		if visit[n - 1] {
			Some(dp[n - 1].0)
		} else {
			None
		}
	};

	let mut ans = 1000.;
	for i in 0..=edges.len() {
		if let Some(r) = calc(i) {
			if r < ans {
				ans = r;
			}
		}
	}
	println!("{}", ans);
}