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>

using namespace std;

template <class T>inline T updmax(T& a, T b) {return a = max(a, b);}
template <class T>inline T updmin(T& a, T b) {return a = min(a, b);}

class Solution {
public:
	void solve() {
		int n, m;
		cin >> n >> m;
		vector<vector<int>> a(n);
		for(int i=0; i < m; i++){
			int s, t;
			cin >> s >> t;
			s--, t--;
			a[s].emplace_back(t);
		}
		vector<double> e(n, 0), p(n, 0);
		for(int i=n-2; i >= 0; i--){
			for(auto x: a[i])e[i] += e[x] + 1;
			e[i] /= a[i].size();
		}
		p[0] = 1.0;
		for(int i=0; i < n; i++){
			for(auto x: a[i]){
				p[x] += p[i]/a[i].size();
			}
		}
		double d = 0.0;
		for(int i=0; i < n; i++){
			if(a[i].size()==1)continue;
			double s = 0.0;
			double f = 0.0;
			for(auto x: a[i]){
				s += e[x];
				updmax(f, e[x]);
			}
			updmax(d, p[i] * (-s+a[i].size()*f)/(a[i].size()*(a[i].size()-1)));
		}
		cout << setprecision(12) <<e[0]-d << endl;
		return;
	};
};

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(nullptr);
	Solution solution;
	solution.solve();
	return 0;
}