Problem:
Score : 600 points

Problem Statement

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}.

Sample Input 1

4 6
1 4
2 3
1 3
1 2
3 4
2 4

Sample Output 1

1.5000000000

If Aoki blocks the passage from Room 1 to Room 2, Takahashi will go along the path 1 → 3 → 4 with probability \frac{1}{2} and 1 → 4 with probability \frac{1}{2}. E = 1.5 here, and this is the minimum possible value of E.

Sample Input 2

3 2
1 2
2 3

Sample Output 2

2.0000000000

Blocking any one passage makes Takahashi unable to reach Room N, so Aoki cannot block a passage.

Sample Input 3

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

Sample Output 3

3.0133333333

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

//#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
//typedef long long ll;

#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define REP(i,a) FOR(i,0,a)
using namespace std;
const int MAX_N=600;
int N,M;
vector<int> G[MAX_N];
double dp[MAX_N];
int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	scanf("%d %d",&N,&M);
	REP(i,M){
		int s,t;
		scanf("%d %d",&s,&t);
		s--;
		t--;
		G[s].push_back(t);
	}
	dp[N-1]=0;
	double ans=N;
	REP(k,N){
		for(int i=N-2;i>=0;i--){
			double mdp=0;
			dp[i]=0;
			for(auto e:G[i]){
				dp[i]+=dp[e];
				mdp=max(mdp,dp[e]);
			}
			if(i==k&&G[i].size()>=2){
				dp[i]-=mdp;
				dp[i]/=G[i].size()-1;
			}else{
				dp[i]/=G[i].size();
			}
			dp[i]++;
		}
		ans=min(ans,dp[0]);
	}
	printf("%.8f\n",ans);
}