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 rep(i,n) for (int i = 0; i < (n); ++i)
#define repit(it, li) for(auto it=li.begin(); it!=li.end(); it++)
using namespace std;
using ll = long long;
using P = pair<int,int>;

int main(){
	int n, m;
	cin>>n>>m;
	vector<int> e[600];
	rep(i, m){
		int s, t;
		cin>>s>>t; s--; t--;
		e[s].push_back(t);
	}

	double dp[600][600];
	// rep(i, n) rep(j, n) dp[i][j]=0;
	rep(j, n) dp[n-1][j]=0;
	for(int i=n-2; i>=0; i--){
		int ne=e[i].size();
		rep(j, n){
			if(i==j && ne>=2){
				double mx=0, sm=0;
				for(int u : e[i]){
					double b=(double)1/(ne-1)*dp[u][j];
					mx=max(mx, b);
					sm+=b;
				}
				dp[i][j]=sm-mx;
			}
			else for(int u : e[i]) dp[i][j]+=(double)1/ne*dp[u][j];
			dp[i][j]+=1;
		}
	}

	double mn=601;
	rep(j, n) mn=min(mn, dp[0][j]);

	cout<<std::setprecision(15)<<mn<<endl;
	return 0;
}