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;

int main() {
	long long N, M;
	double ANS = 0.0;
	vector<vector<long long> > V;
	vector<double> X;
	vector<vector<pair<double, long long> > > VP;
	cin >> N >> M;
	V.resize(N);
	X.resize(N);
	VP.resize(N);
	for (int i = 0; i < M; i++) {
		long long S, T;
		cin >> S >> T;
		S--, T--;
		V[S].push_back(T);
	}
	X[N - 1] = 0.0;
	for (int i = N - 2; i >= 0; i--) {
		double COUNT = 0.0;
		for (long long next : V[i]) {
			COUNT += X[next];
			VP[i].push_back({ X[next], next });
		}
		X[i] = 1.0 + COUNT / V[i].size();
	}
	ANS = X[0];
	for (int i = 0; i < N; i++) {
		if (VP[i].size() <= 1) continue;
		sort(VP[i].begin(), VP[i].end(), greater<pair<double, long long> >());
		long long max = VP[i][0].second;
		vector<double> Y;
		Y.assign(N, 0.0);
		for (int j = N - 2; j >= 0; j--) {
			double COUNT = 0.0;
			for (long long next : V[j]) {
				if (j == i && next == max) continue;
				COUNT += Y[next];
			}
			Y[j] = 1.0 + COUNT / (V[j].size() - (i == j));
		}
		ANS = min(ANS, Y[0]);
	}
	printf("%.12lf\n", ANS);
}