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)

#include<bits/stdc++.h>
using namespace std;
 
#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
#define repl(i,s,n) for(int i=s; i<=n; ++i)
#define rer(i,l,u) for(int (i)=(int)(l);(i)<=(int)(u);++(i))
#define reu(i,l,u) for(int (i)=(int)(l);(i)<(int)(u);++(i))
#if defined(_MSC_VER) || __cplusplus > 199711L
#define aut(r,v) auto r = (v)
#else
#define aut(r,v) __typeof(v) r = (v)
#endif
#define each(it,o) for(aut(it, (o).begin()); it != (o).end(); ++ it)
#define ktya(x) sort(all(x))
#define maxs(x,y) (x = max(x,y))
#define mins(x,y) (x = min(x,y))
#define all(o) (o).begin(), (o).end()
#define pb(x) push_back(x)
#define mp(x,y) make_pair((x),(y))
#define mset(m,v) memset(m,v,sizeof(m))
#define INF 1e18
#define INFLL 1000000000000000007LL
#define SIZE 200105
#define REP(i,n) for(int i=0;i<(int)n;++i)
#define FOR(i,c) for(__typeof((c).begin())i=(c).begin();i!=(c).end();++i)
#define ALL(c) (c).begin(), (c).end()
#define mind(a,b) (a>b?b:a)
#define maxd(a,b) (a>b?a:b)
#define PI (acos(-1))
typedef vector<int> vi; typedef pair<int, int> pii; typedef vector<pair<int, int> > vpii; typedef long long ll;
typedef pair<int,ll> pill; typedef pair<ll,int> plli; 
typedef pair<double, int> pdi;
template<typename T, typename U> inline void amin(T &x, U y) { if(y < x) x = y; }
template<typename T, typename U> inline void amax(T &x, U y) { if(x < y) x = y; }
typedef complex<double> P;
typedef pair<pill,int> PA;
//ll MOD;
ll MOD = 1000000007;
// ll MOD=998244353;
typedef ll Weight;
struct Edge {
	int src, dst;
	ll weight;
};
int N,M;
double dp[200000];
double dp2[200000];
vi gr[700];
int main(){
	cin>>N>>M;
	rep(i,M){
		int s,t;
		cin>>s>>t;s--;t--;
		gr[s].pb(t);
	}
	dp[N-1]=0.0;
	for(int i=N-2;i>=0;i--){
		dp[i]=0.0;
		int num=gr[i].size();
		rep(j,num){
			int x=gr[i][j];
			dp[i]+=(1.0/(double)num)*(dp[x]+1.0);
		}
	}
	double ans=dp[0];
	//cout<<dp[0]<<endl;
	for(int j=N-2;j>=0;j--){
		int check=j;
		bool flag=true;
		for(int i=N-2;i>=0;i--){
			if(i==check){
				dp2[i]=0.0;
				int num=gr[i].size();
				if(num==1)flag=false;
				else{
					double ma=dp2[gr[i][0]];
					rep(j,num){
						int x=gr[i][j];
						ma=max(ma,dp[x]);
						dp2[i]+=(1.0/(double)(num-1))*(dp2[x]+1.0);
					}
					dp2[i]-=(ma+1.0)/(double)(num-1);
				}

			}else{
				dp2[i]=0.0;
				int num=gr[i].size();
				rep(j,num){
					int x=gr[i][j];
					dp2[i]+=(1.0/(double)num)*(dp2[x]+1.0);
				}
			}
		}
		if(flag)ans=min(ans,dp2[0]);
	}
	cout<< setprecision(10)<<ans<<endl;

}