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 <cstdio>
#include <algorithm>
#include <vector>
using namespace std;
int N;
vector<int> adjlist[599];
double dptable[600];
double dp(int v) {
	if (dptable[v] > -0.5) return dptable[v];
	if (v == N - 1) return dptable[v] = 0;
	dptable[v] = 0;
	for (int i: adjlist[v]) dptable[v] += dp(i) / adjlist[v].size();
	return ++dptable[v];
}
int main() {
	int M, s, t;
	scanf("%d %d", &N, &M);
	for (int i = 0; i < M; ++i) {
		scanf("%d %d", &s, &t);
		adjlist[s - 1].push_back(t - 1);
	}
	for (int i = 0; i < N; ++i) dptable[i] = -1;
	double ans = dp(0);
	for (int i = 0; i < N - 1; ++i) {
		for (int j = 0; j < N; ++j) dptable[j] = -1;
		if (adjlist[i].size() > 1) {
			double temp = 0, maxval = 0;
			for (int j: adjlist[i]) {
				temp += dp(j);
				maxval = max(maxval, dp(j));
			}
			temp -= maxval;
			temp /= (adjlist[i].size() - 1);
			dptable[i] = temp + 1;
			ans = min(ans, dp(0));
		}
	}
	printf("%.9lf", ans);
	return 0;
}