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>
typedef long long ll;
using namespace std;
const ll mod = 1000000007;
const ll mod998 = 998244353;
const ll intmax = 2147483647;
const ll llmax = 9223372036854775807;
const char sp = ' ';

int N, M, s, t;
double ex[600], res;
vector<int>v[600];

int main() {
	cin >> N >> M;
	for (int i = 0; i < M; i++) {
		cin >> s >> t;
		v[s - 1].push_back(t - 1);
	}
	for (int i = N - 2; i >= 0; i--) {
		for (int j : v[i]) {
			ex[i] += ex[j];
		}
		ex[i] /= v[i].size();
		ex[i] += 1;
	}
	res = ex[0];
	for (int i = 0; i < N - 1; i++) {
		if (v[i].size() == 1)continue;
		double EX[600];
		for (int j = i + 1; j < N; j++) {
			EX[j] = ex[j];
		}
		EX[i] = 0;
		double mx = 0;
		for (int j : v[i]) {
			EX[i] += EX[j];
			mx = max(mx, EX[j]);
		}
		EX[i] -= mx;
		EX[i] /= v[i].size() - 1;
		EX[i] += 1;
		for (int j = i - 1; j >= 0; j--) {
			EX[j] = 0;
			for (int k : v[j]) {
				EX[j] += EX[k];
			}
			EX[j] /= v[j].size();
			EX[j] += 1;
		}
		res = min(res, EX[0]);
	}
	printf("%.10f\n", res);
}