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: Wrong Answer)

#include <bits/stdc++.h>
#define N 605

using namespace std;

int n, m;
int deg[N], ideg[N], Outdeg[N];
bool G[N][N], iG[N][N];
double f[N], p[N], outp[N];

inline void cmin(double &u, double v) { if(u > v) u = v; }
inline void cmax(double &u, double v) { if(u < v) u = v; }

void build() {
	int u, v;
	cin >> n >> m;
	for(int i = 1; i <= m; ++i) {
		cin >> u >> v;
		G[u][v] = true, ++deg[v];
		iG[v][u] = true, ++ideg[u];
	}
	for(int i = 1; i <= n; ++i) {
		Outdeg[i] = ideg[i];
		if(Outdeg[i]) outp[i] = 1.0 / Outdeg[i];
	}
	return;
}

void topo_1() {
	queue<int> que;
	que.push(n);
	while(!que.empty()) {
		int u = que.front(); que.pop();
		for(int v = 1; v <= n; ++v) {
			if(!iG[u][v]) continue;
			f[v] += (1.0 + f[u]) * outp[v];
			if((--ideg[v]) == 0) que.push(v);
		}
	}
	return;
}
void topo_2() {
	queue<int> que;
	que.push(1), p[1] = 1.0;
	while(!que.empty()) {
		int u = que.front(); que.pop();
		for(int v = 1; v <= n; ++v) {
			if(!G[u][v]) continue;
			p[v] += p[u] * outp[u];
			if((--deg[v]) == 0) que.push(v);
		}
	}
	return;
}

int main() {
	ios :: sync_with_stdio(false);
	build();
	topo_1(), topo_2();
	double ans = f[1], res;
	double p_1, p_2, maxf;
	for(int i = 1; i <= n; ++i) {
		if(Outdeg[i] <= 1) continue;
		res = maxf = 0;
		p_1 = outp[i], p_2 = 1.0 / (Outdeg[i] - 1);
		for(int j = 1; j <= n; ++j) {
			if(!G[i][j]) continue;
			res += 1.0 + f[j];
			cmax(maxf, 1.0 + f[j]);
		}
		cmin(ans, f[1] + p[i] * ((res - maxf) * p_2 - res * p_1));
	}
	cout << fixed << setprecision(7) << ans << endl;
	return 0;
}