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 SZ(x) (int)(x.size())

using ll = long long;
using ld = long double;
using P = pair<int, int>;
using vi = vector<int>;
using vvi = vector<vector<int>>;
using vll = vector<ll>;
using vvll = vector<vector<ll>>;
const ld eps = 1e-9;
const ll MOD = 1000000007;
const int INF = 1000000000;
const ll LINF = 1ll<<50;

template<typename T>
void printv(const vector<T>& s) {
  for(int i=0;i<(int)(s.size());++i) {
    cout << s[i];
    if(i == (int)(s.size())-1) cout << endl;
    else cout << " ";
  }
}

template<typename T1, typename T2>
ostream& operator<<(ostream &os, const pair<T1, T2> p) {
  os << p.first << ":" << p.second;
  return os;
}

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(10);

  int n, m; cin >> n >> m;
  vvi g(n);
  vi d(n);
  for(int i=0;i<m;++i) {
    int s, t; cin >> s >> t;
    s--; t--;
    g[s].push_back(t);
    d[s]++;
  }
  
  vector<double> dp(n, 0.0);
  vector<double> dpma(n, 0.0);
  dp[n-1] = 0.0;
  for(int i=n-2;i>=0;--i) {
    double su = 0.0;
    double ma = 0.0;
    for(int j=0;j<SZ(g[i]);++j) {
      su += dp[g[i][j]] + 1;
      ma = max(ma, dp[g[i][j]] + 1);
    }
    dp[i] = su / d[i];
    dpma[i] = ma;
  }
  //printv(dp);

  double ans = INF;
  for(int i=0;i<n-1;++i) {
    vector<double> dp2(n);
    for(int j=i+1;j<n;++j) {
      dp2[j] = dp[j];
    }
    if(d[i] > 1) dp2[i] = (dp[i] * d[i] - dpma[i]) / (d[i] - 1);
    else dp2[i] = dp[i];
    for(int j=i-1;j>=0;--j) {
      double su = 0;
      for(int k=0;k<SZ(g[j]);++k) {
        int nxt = g[j][k];
        su += dp2[nxt] + 1;
      }
      dp2[j] = su / d[j];
    }
    //printv(dp2);
    ans = min(ans, dp2[0]);
  }
  cout << ans << endl;

}