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 <iostream>
#include <fstream>
#include <stdio.h>
#include <vector>
#include <string>
#include <math.h>
#include <algorithm>
#include <map>
#include <set>

#define sz(x) ((int)x.size())
#define all(x) (x).begin(), (x).end()
#define pb(x) push_back(x)
#define mp(x, y) make_pair(x, y)

typedef long long int64;

using namespace std;

int main() {
	ios_base::sync_with_stdio(0);
    
	int n, m;
	cin >> n >> m;
    vector<vector<int>> g(n);
	vector<pair<int, int>> a;
	for (int i = 0; i < m; ++i) {
	    int x, y;
		cin >> x >> y;
		--x;
		--y;
		g[x].push_back(y);
	}
	double inf = 10000000;
	vector<double> dp0(n, 0);
	vector<double> dp1(n, inf);
	for (int x = n - 2; x >= 0; --x) {
	    for (int y : g[x]) {
		    dp0[x] += dp0[y];
		}
		dp0[x] /= sz(g[x]);
		dp0[x] += 1;
		
		double sum = 0;
		double min_val = inf;
		for (int y : g[x]) {
		    dp1[x] += dp0[y];
	    	min_val = min(min_val, dp1[y] - dp0[y]);
		}
		dp1[x] += min_val;
		dp1[x] /= sz(g[x]);
		dp1[x] += 1;
		if (sz(g[x]) <= 1) {
			continue;
		}
		double max_val = -inf;
		sum = 0;
		for (int y : g[x]) {
		    sum += dp0[y];
			max_val = max(max_val, dp0[y]);
		}
		double add = 1 + (sum - max_val) / (sz(g[x]) - 1);
		dp1[x] = min(dp1[x], add);
	}
	cout.precision(10);
	cout << min(dp0[0], dp1[0]) << endl;
	return 0;
}