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;
typedef long long ll;
const ll MOD = 1e9 + 7;
const ll INF = 1LL << 60;
double dp[600];
double A[600], B[600];
vector<int> G[600];
int main() {
	int N, M;
	cin >> N >> M;
	for (int i = 0; i < M; i++) {
		int s, t;
		cin >> s >> t;
		s--; t--;
		G[s].push_back(t);
	}
	dp[N - 1] = 0;
	for (int v = N - 2; v >= 0; v--) {
		for (int to : G[v]) {
			dp[v] += (dp[to] + 1);
		}
		dp[v] /= G[v].size();
	}
	double ans = dp[0];
	for (int s = 0; s < N; s++) {
		if (G[s].size() == 1) continue;
		A[s] = 1;
		B[s] = 0;
		for (int i = s + 1; i < N; i++) {
			A[i] = 0;
			B[i] = dp[i];
		}
		for (int i = s - 1; i >= 0; i--) {
			A[i] = B[i] = 0;
			for (int to : G[i]) {
				A[i] += A[to] / G[i].size();
				B[i] += (B[to] + 1) / G[i].size();
			}
		}
		for (int to : G[s]) {
			double x = (dp[s] * G[s].size() - (dp[to] + 1)) / (G[s].size() - 1);
			ans = min(ans, A[0] * x + B[0]);
		}
	}
	printf("%.15lf\n", ans);
}