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 inf = 1e9;
const ll maxn = 610;
int a[maxn*maxn], b[maxn*maxn];
double f[maxn], ans;
vector <int> vec[maxn];
ll n, m;
double DP(double p) {
	for(int i = 1; i <= n; ++i) {
		f[i] = 0;
	}
	f[n] = 0;
	for(int i = n - 1; i >= 1; --i){
		ll len = vec[i].size();
		if(p == i && len == 1) {
			f[i] = inf;
			continue;
		}
		double t = 1.0 / (len - (p == i));
		double mx = 0;
		for(int j = 0; j < len; ++j) {
			int v = vec[i][j];
			mx = max(mx, (f[v]+1)*t);
			f[i] += (f[v] + 1) * t;
		}
		if(p == i)	f[i] -= mx;
	}
	return f[1];
}
int main () {
	cin >> n >> m;
	for(int i = 1; i <= m; ++i) {
		cin >> a[i] >> b[i];
		vec[a[i]].push_back(b[i]);
	}
	ans = DP(0);
	for(int i = 1; i <= n; ++i) {
		ans = min(ans, DP(i));
	}
	printf("%.9f\n", ans);
}