Problem: 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}.

Examples

Input

4 6
1 4
2 3
1 3
1 2
3 4
2 4


Output

1.5000000000


Input

3 2
1 2
2 3


Output

2.0000000000


Input

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


Output

3.0133333333

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

int main()
{
	using vd = std::vector<long double>;
	int N, M;
	std::vector<std::vector<int>> edges;
	scanf("%d%d", &N, &M);
	edges.resize(N);
	for (int i{}; i < M; i++)
	{
		int s, t;
		scanf("%d%d", &s, &t);
		s--; t--;
		edges[s].push_back(t);
	}
	vd E(N);
	for (int i{N - 2}; i >= 0; i--)
		for (auto& e: edges[i])
			E[i] += (E[e] + 1) / edges[i].size();
	long double ans{E[0]};
	for (int i{N - 2}; i >= 0; i--)
	{
		if (edges[i].size() == 1) continue;
		long double max{};
		for (auto& e: edges[i])
			max = std::max(max, E[e] + 1);
		vd ignoreE(N);
		for (int j{N - 2}; j >= 0; j--)
		{
			for (auto& e: edges[j])
				ignoreE[j] += ignoreE[e] + 1;
			if (j == i)
				ignoreE[j] -= max;
			int size{(int)edges[j].size()};
			if (j == i) size--;
			ignoreE[j] /= size;
		}
		ans = std::min(ans, ignoreE[0]);
	}
	printf("%.9Lf\n", ans);

	return 0;
}