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 <iostream>
#include <vector>
#include <algorithm>
#include <iomanip>

int main() {
	int n, m;
	std::cin >> n >> m;

	std::vector<std::vector<int>> edges(n);
	for (int i = 0; i < m; ++i) {
		int s, t;
		std::cin >> s >> t;
		--s, --t;

		edges[s].push_back(t);
	}

	for (auto&& vec : edges)
		std::sort(vec.begin(), vec.end());

	double ans = n + 1;
	
	std::vector<double> dp(n);
	for (int i = 0; i < n - 1; ++i) {
		std::fill(dp.begin(), dp.end(), 0);

		for (int j = n - 2; j >= 0; --j) {
			auto&& es = edges[j];
			if (es.size() == 1)
				dp[j] = dp[es[0]] + 1;
			else {
				double max = 0;
				int maxIdx = -1;
				for (int k = 0; k < es.size(); ++k) {
					if (dp[es[k]] > max) {
						max = dp[es[k]];
						maxIdx = k;
					}
				}
				for (int k = 0; k < es.size(); ++k)
					if (i != j || k != maxIdx)
						dp[j] += (dp[es[k]] + 1) / (es.size() - (i == j));
			}
		}

		if (dp[0] < ans)
			ans = dp[0];
	}

	std::cout << std::setprecision(10);
	std::cout << ans << std::endl;
}