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: Wrong Answer)

#include<bits/stdc++.h>
#define FAST ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define dofloat cout<<fixed<<setprecision(8)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define bitcount __builtin_popcount
#define all(vec) vec.begin(),vec.end()
#define rall(vec) vec.rbegin(),vec.rend()
using namespace std;
typedef long long ll;
typedef long double ld;
typedef vector<long long> vl;
typedef pair<long long,long long> pll;
typedef vector<pair<long long,long long> > vll;
typedef vector<pair<int,int> > vii;
typedef vector<int> vi;
typedef pair<int,int> ii;
const long long MOD=1000000007;
const long long MAX=100005;
const long double PI=3.14159265359;
const long double G=9.807;
const long long INF=1e18;
const long double EPS=1e-6;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
bool isprime(ll a){
  if(a==2)return 1;
  if(!(a&1))return 0;
  for(ll i=3;i*i<=a;i+=2)
    if(a%i==0)return 0;
  return 1;
}
ll mpow(ll base,ll exponent,ll modulus){
  if(modulus==1)return 0;
  long long result = 1;
  base=base%modulus;
  while(exponent){
    if(exponent%2 == 1)
	  result=(result*base)%modulus;
	  exponent=exponent>>1;
	  base=(base*base)%modulus;
  }
  return result;
}
ll minv(ll a,ll mod){
  ll _gcd=gcd(a,mod);
  assert(_gcd==1);
  return mpow(a,mod-2,mod);
}


/*
ll ncr(ll N,ll K){
	if(N<K)return 0;
	if(K==0)return 1;
	if(N==0)return 0;

	ll den=1;
	for(ll i=1;i<=K;i++)den*=i;
	ll num=1;
	while(K--){
		num*=N;
		if(num%den==0){
			num/=den;
			den=1;
		}
		N--;
	}

	return num;
}
*/
ll n,m;
ld E[605];
pll e[1000000];
vl g[605];
int main(){FAST;
cin>>n>>m;
for(int i=1;i<=m;i++){
	ll u,v;cin>>u>>v;
	g[u].pb(v);
	e[i] = {u,v};
}
ld ans = n-1;
for(ll i=1;i<=m;i++){
	ld EE = 0;
	for(ll j=1;j<=n;j++)E[j]=0;
	for(ll j=n-1;j>=1;j--){
		ld tot = 0;
		ll sz = g[j].size();
		for(auto it:g[j]){
			if(mp(j,it)==e[i]){
				sz--;
				continue;
			}
			tot+=E[it]+1;
		}
		if(!sz)E[j] = 0;
		else E[j] = tot/sz;
	}
	ans=min(ans,E[1]);
}
cout<<ans;
return 0;
}