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<iomanip>
#include<float.h>
using namespace std;
using uint = unsigned int;

int main() {
	uint N, M;
	cin >> N >> M; 
	
	vector<vector<uint>> nodes(N);
	uint s, t;
	
	for(uint m=0; m<M; m++) {
		cin >> s >> t;
		nodes[s-1].push_back(t-1);
	}
	
	double ans = DBL_MAX;
	
	for(uint j=0; j<N; j++) {
		vector<double> e_dists(N);
		for(int i=N-2; i>=0; i--) {
			double max_dist = 0;
			double n_dist = 0;
			
			for(uint k=0; k<nodes[i].size(); k++) {
				n_dist += e_dists[nodes[i][k]];
				max_dist = max(max_dist, e_dists[nodes[i][k]]);
			}
			
			if ((nodes[i].size() > 1) && (i == j)) {
				e_dists[i] = 1 + (n_dist - max_dist) / (nodes[i].size() - 1);
			} else {
				e_dists[i] = 1 + n_dist / nodes[i].size();
			}
		}
		ans = min(ans, e_dists[0]);
	}
	
	cout << std::setprecision(7) << ans;
	
	return 0;
}