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() {
	int N, M;
	cin >> N >> M;
	vector<vector<int>>to(N);
	for (int i = 0; i < M; ++i) {
		int s, t;
		cin >> s >> t;
		s--;
		t--;
		to[s].push_back(t);
	}
	vector<double>dp(N);
	dp[N - 1] = 0;
	for (int i = N - 2; i >= 0; --i) {
		double  now = 0;
		for (auto j : to[i]) {
			now += dp[j];
		}
		now /= to[i].size();
		now += 1;
		dp[i] = now;
	}
	double ans = dp[0];
	
	for (int i = 0;i<N-1;++i) {
		if (1 == to[i].size()) {
			continue;
		}
		int nu = to[i][0];
		for (auto j : to[i]) {
			if (dp[nu] < dp[j]) {
				nu = j;
			}
		}
		vector<double>dp0(N);
		dp0[N - 1] = 0;
		for (int j = N - 2; j >= 0; --j) {
			double  now = 0;
			for (auto k : to[j]) {
				if (j == i && k == nu) {
					continue;
				}
				now += dp0[k];
			}
			int deg = to[j].size();
			if (j == i) {
				deg--;
			}
			now /= deg;
			now += 1;
			dp0[j] = now;
		}
		ans = min(ans, dp0[0]);
	}
	printf("%.10f\n", ans);
	return 0;
}