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>
using namespace std;

double prob[610], expe[610], esum[610], mx[610];

int main() {
	int N, M;
	cin >> N >> M;
	vector<vector<int>> v(N+1);
	for(int i=0; i<M; ++i){
		int s, t;
		cin >> s >> t;
		v[s].emplace_back(t);
	}
	prob[1]=1.0;
	for(int i=1; i<N; ++i){
		for(int j=0; j<v[i].size(); ++j){
			prob[v[i][j]] += prob[i]/v[i].size();
		}
	}
	for(int i=N-1; i>0; --i){
		for(int j=0; j<v[i].size(); ++j){
			esum[i] += expe[v[i][j]];
			mx[i]=max(mx[i], expe[v[i][j]]);
		}
		expe[i]=1.0+esum[i]/v[i].size();
	}
	double ans=0.0;
	for(int i=1; i<N; ++i){
		if(v[i].size()>1){
			ans=max(ans, prob[i]*(expe[i]-(esum[i]-mx[i])/(v[i].size()-1)-1.0));
		}
	}
	cout << setprecision(15) << expe[1]-ans << endl;
	return 0;
}