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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <iomanip>
#define REP(i, a, b) for (auto i = (a); i < (b); ++i)
using namespace std;

const int inf = 1e9;
const int N = 600;
bool adj[N][N];
int deg[N];
double E[N][2];

signed main() {
	ios_base::sync_with_stdio(false);
	cin.tie(nullptr);

	int n, m; cin >> n >> m;

	while(m--) {
		int s, t; cin >> s >> t; --s, --t;
		adj[s][t] = 1;
		++deg[s];
	}

	auto solve = [&](bool b) {
		REP(k, 1, n) {
			int i = n - k - 1;
			E[i][b] = 1;
			REP(j, i + 1, n - 1)
				if (adj[i][j])
					E[i][b] += E[j][b] / deg[i];
		}
	};

	solve(0);
	double ans = E[0][0];

	REP(i, 0, n - 1) {
		if (deg[i] == 1) continue;

		int mxpos; double mx = -1;
		REP(j, i + 1, n) {
			if (adj[i][j] && mx < E[j][0]) {
				mxpos = j;
				mx = E[j][0];
			}
		}

		--deg[i]; adj[i][mxpos] = 0;
		solve(1);
		++deg[i]; adj[i][mxpos] = 1;

		ans = min(ans, E[0][1]);
	}

	cout << fixed << setprecision(10) << ans;
}