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 rep(i, n) for(int i=0; i<n; i++)
#define pb push_back


int main()
{
	int n, m;
	double p[610];
	vector<int> ed[610], inv_ed[610];
	cin >> n >> m;
	int s, t;
	rep(i, m)
	{
		cin >> s >> t;
		s--; t--;
		ed[s].pb(t);
		inv_ed[t].pb(s);
	}
	p[n-1]=0;
	for(int i=n-2; i>=0; i--)
	{
		p[i]=1;
		rep(j, ed[i].size())
		{
			p[i]+=p[ed[i][j]]/(double)ed[i].size();
		}
	}
	int mx[610];
	rep(i, n-2)
	{
		mx[i]=ed[i][0];
		rep(j, ed[i].size())
		{
			if(p[ed[i][j]]>p[mx[i]]) mx[i]=ed[i][j];
		}
	}
	double ans=p[0];
	rep(k, n-2)
	{
		if(ed[k].size()>1)
		{
			for(int i=n-2; i>=0; i--)
			{
				p[i]=0;
				rep(j, ed[i].size())
				{
					if(i!=k || ed[i][j]!=mx[i]) p[i]+=p[ed[i][j]]/(double)ed[i].size();
				}
				if(i==k) p[i]=p[i]*(double)ed[i].size()/(double)(ed[i].size()-1);
				p[i]+=1.0;
			}
			ans=min(ans, p[0]);
		}
	}
	cout << setprecision(9) << fixed << ans << endl;
	return 0;
}