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;
#define all(V) V.begin(),V.end()
#define ll long long
#define MOD 1000000007

vector<int> G[610];
pair<int, int> E[200000];
double cost[610], les[610];

int main() {
	int N, M, s, t, sz;
	cin >> N >> M;
	for (int i = 0;i < M;i++) {
		cin >> s >> t;
		G[s].push_back(t);
		E[i] = { s,t };
	}
	for (int i = N;i > 0;i--) {
		cost[i] = 1.0;
		//(接続先の期待値+1)*(そこに進む確率)
		for (int j = i - 1;j > 0;j--) {
			sz = G[j].size();
			for (int k : G[j]) {
				cost[j] += cost[k] / sz;
			}
		}
		les[i] = cost[1];
		for (int j = 1;j <= N;j++)cost[j] = 0.0;
	}

	for (int i = N;i > 0;i--) {
		sz = G[i].size();
		for (int k : G[i]) {
			cost[i] += (cost[k] + 1) / sz;
		}
	}

	double ans = 0, tp;
	for (int i = 0;i < M;i++) {
		tie(s, t) = E[i];
		sz = G[s].size();
		if (sz == 1)continue;
		tp = cost[s] - (cost[t] + 1.0) / sz;
		tp *= (double)sz / (sz - 1);
		tp = (cost[s] - tp) * les[s];
		ans = max(tp, ans);
	}
	printf("%.10f", cost[1] - ans);
}