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 pb push_back
#define mp make_pair
#define int long long
#define inf 1000000007
#define LINF 100000000000000007LL
#define ll long long
using namespace std;
vector<int> nex[1145];
double dp[1145];
signed main(){
	int n,m;
	cin>>n>>m;
	for(int i=0;i<m;i++){
		int s,t;
		cin>>s>>t;
		nex[s].pb(t);
	}
	for(int i=1;i<=n-1;i++){
		sort( nex[i].begin(), nex[i].end() );
	}
	double ans = 100000.0;
	for(int k=1;k<=n-1;k++){
		dp[n]=0.0;
		for(int i=n-1;i>=1;i--){
			int cnt = 0;
			double sum = 0.0;
			if( i == k ){
				vector<double> vec;
				for(int j=0;j<nex[i].size();j++){
					int nx = nex[i][j];
					vec.pb(dp[nx]);
					sum += dp[nx];
					cnt++;
				}
				sort( vec.begin(), vec.end() );
				if( vec.size() > 1 ){
					sum -= vec[vec.size()-1];
					cnt--;
				}
			}
			else{
				for(int j=0;j<nex[i].size();j++){
					int nx = nex[i][j];
					sum += dp[nx];
					cnt++;
				}
			}
			dp[i] = sum/cnt + 1.0;
		}
		if( ans > dp[1] ) ans = dp[1];
	}
	cout<<setprecision(16)<<ans<<endl;
	return 0;
}