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 modulo 998244353
#define mod(mod_x) ((((long long)mod_x+modulo))%modulo)
#define Inf 100000000.0


int main(){
    
	int N,M;
	cin>>N>>M;
	
	vector<vector<int>> E(N,vector<int>());
	
	for(int i=0;i<M;i++){
		int a,b;
		cin>>a>>b;
		a--;b--;
		E[a].push_back(b);
	}
	
	vector<double> dp(N,0.0);
	
	for(int i=N-2;i>=0;i--){
		for(int j=0;j<E[i].size();j++){
			dp[i] += (dp[E[i][j]]+1.0)/E[i].size();
		}
	}
	double ans = dp[0];
	for(int i=0;i<N-1;i++){
		int ind = -1;
		double maxi=0.0;
		if(E[i].size()==1)continue;
		for(int j=0;j<E[i].size();j++){
			if(dp[E[i][j]]>maxi){
				maxi=dp[E[i][j]];
				ind=j;
			}
		}
		
		int to = E[i][ind];
		E[i].erase(E[i].begin()+ind);
		
		vector<double> dp2(N,0.0);
		for(int i=N-2;i>=0;i--){
			for(int j=0;j<E[i].size();j++){
				dp2[i] += (dp2[E[i][j]]+1.0)/E[i].size();
			}
		}
		ans = min(ans,dp2[0]);
		
		E[i].push_back(to);
	}
	
	cout<<fixed<<setprecision(15)<<ans<<endl;
		
	
	
	
    return 0;
}