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>
 
#define ll long long int
#define pb push_back
#define eb emplace_back
#define xx first
#define yy second
#define INF 0x3f3f3f3f

 
using namespace std;
const int N=601, mod= 1e9+7;
vector<int> g[N];
vector<long double> dp1(N),dp2(N);

int main(){
	ios_base::sync_with_stdio(0);	cin.tie(0);
	int n,m;
	
	cin>>n>>m;
	for(int i=0;i<m;i++){
		int u,v;
		cin>>u>>v;
		g[u].pb(v);
	}
	
	for(int i=n-1;i>0;i--){
		for(auto v:g[i]){
			dp1[i]+=dp1[v]+1;
		}
		dp1[i]/=g[i].size();
	}
	
	long double ans=dp1[1];
	
	
	for(int i=1;i<n;i++){
		if(g[i].size()<2) continue;
		
		long double mx=-1;
		for(auto j:g[i]){
			if( mx<dp1[j] ){
				mx=dp1[j];
			}
		}
		
		for(auto k:g[i]){
			if( dp1[k]==mx ){
				
				dp2[n]=0;
				for(int j=n-1;j>0;j--){
					dp2[j]=0;
					int ct=0;
					for(auto v:g[j]){
						if( i==j && v==k ) continue;
						ct++;
						dp2[j]+=dp2[v]+1;
					}
					dp2[j]/=ct;
				}
				ans=min(ans,dp2[1]);
				
			}
		}
		
		
	}
	
	
	cout<<fixed<<setprecision(8)<<ans<<"\n";
	
}