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<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<complex>
#include<bitset>
#include<stack>
#include<unordered_map>
#include<utility>
#include<tuple>
using namespace std;
typedef long long ll;
typedef unsigned int ui;
const ll mod = 1000000007;
const ll INF = (ll)1000000007 * 1000000007;
typedef pair<int, int> P;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define Per(i,sta,n) for(int i=n-1;i>=sta;i--)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
typedef long double ld;
const ld eps = 1e-8;
const ld pi = acos(-1.0);
typedef pair<ll, ll> LP;
int dx[4]={1,-1,0,0};
int dy[4]={0,0,1,-1};

int n,m;
vector<int> G[610];
ld e[610];
int mi[610];


void solve(){
    cin >> n >> m;
    rep(i,m){
        int a,b;cin >> a >> b;a--;b--;
        G[a].push_back(b);
    }
    per(i,n-1){
        ld siz=G[i].size();
        mi[i]=-1;
        for(int t:G[i]){
            e[i]+=e[t]/siz;
            if(mi[i]==-1){
                mi[i]=t;
            }
            else if(e[mi[i]]<e[t]){
                mi[i]=t;
            }
        }
        e[i]+=1;
        //cout << i << " " << mi[i] << " " << e[i] << endl;
    }
    ld ans=e[0];
    rep(k,n-1){
        vector<ld> e2(n,0);
        bool flag=true;
        per(i,n-1){
            ld siz=G[i].size();
            if(k==i) siz-=1;
            if(abs(siz)<eps) {
                flag=false;
                break;
            };
            for(int t:G[i]){
                if(k==i && t==mi[i]) continue;
                e2[i]+=e2[t]/siz;
            }
            e2[i]+=1;
        } 
        if(flag)ans=min(ans,e2[0]);
    }
    cout << ans << endl;
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout << fixed << setprecision(50);
    solve();
}