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;
#define rep(i,n) for(ll i = 0;i<((ll)(n));i++)
#define reg(i,a,b) for(ll i = ((ll)(a));i<=((ll)(b));i++)
#define irep(i,n) for(ll i = ((ll)(n)-1);i>=0;i--)
#define ireg(i,a,b) for(ll i = ((ll)(b));i>=((ll)(a));i--)
template<class T = int> using V = vector<T>;
template<class T = int> using VV = V< V<T> >;
template<class T = int> void print(vector<T> v){for(auto a:v)cout<<a<<" ";cout<<endl;}

/*
DAGになってる！！ 
コスト求めるのがO(m)でいける
*/

ll n,m;
double ans=1e9;
V<ll> v[601];

int main(void){
	cin>>n>>m;
	rep(i,m){
		ll s,t;
		cin>>s>>t;
		v[s].push_back(t);
	}
	reg(i,1,n){
		double d[601]={};
		ireg(j,1,n-1){
			int ma = 0;
			rep(k,v[j].size()){
				if(d[v[j][k]]>d[v[j][ma]])ma=k;
				d[j]+=d[v[j][k]];
			}
			if(i==j && v[j].size()>1){
				d[j]-=d[v[j][ma]];
				d[j]/=v[j].size()-1;
			}else{
				d[j]/=v[j].size();
			}
			d[j]+=1;
		}
		// reg(j,1,n)cerr<<d[j]<<" ";
		// cerr<<endl;
		ans=min(ans,d[1]);
	}
	cout<<fixed<<setprecision(15)<<ans<<endl;
	return 0;
}