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: Time Limit Exceeded)

#include"bits/stdc++.h"
#include<boost/multi_array.hpp>
#include<boost/optional.hpp>
#include<boost/range/irange.hpp>
#include<boost/range/algorithm.hpp>
#include<boost/range/adaptors.hpp>
#include<boost/operators.hpp>
#include<boost/dynamic_bitset.hpp>

namespace adaptor = boost::adaptors;
using index_t = std::size_t;

void Main()
{
	int N, M;
	std::cin >> N >> M;
	std::vector<std::pair<std::set<int>, std::vector<std::pair<int, int>>>> edge(N);
	for (auto i : boost::irange(0, M))
	{
		int s, t;
		std::cin >> s >> t;
		--s; --t;
		edge[s].first.emplace(i);
		edge[s].second.emplace_back(i, t);
	}
	double min = std::numeric_limits<double>::max();
	for (auto s : boost::irange(-1, M))
	{
		bool flag = false;
		boost::multi_array<double, 2> dp(boost::extents[N][N]);
		dp[0][0] = 1;
		for (auto i : boost::irange(0, N - 1))
		{
			auto size = edge[i].first.size() - edge[i].first.count(s);
			if (size == 0)
			{
				flag = true;
				break;
			}
			auto d = 1.0 / size;
			for (auto p : edge[i].second)
			{
				if (p.first == s)
				{
					continue;
				}
				for (auto j : boost::irange(0, N - 1))
				{
					dp[p.second][j + 1] += d * dp[i][j];
				}
			}
		}
		if (!flag)
		{
			double E = 0;
			for (auto i : boost::irange(1, N))
			{
				E += i * dp[N - 1][i];
			}
			min = std::min(min, E);
		}
	}
	std::cout << min << std::endl;
}

int main()
{
	std::cin.tie(nullptr);
	std::ios_base::sync_with_stdio(false);
	std::cout << std::fixed << std::setprecision(15);
	Main();
}