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>
#define rep(i, n) for (int i = 0; i < (n); ++i)
using namespace std;
typedef long long ll;

int main() {
  double N, M;
  cin >> N >> M;
  vector<vector <int> > to(N);
  rep(i, M) {
    int s, t;
    cin >> s >> t;
    --s; --t;
    to[s].push_back(t);
  }
  vector<double> DP(N);
  for (int v = N - 2; v >= 0; --v) {
    double now = 0;
    for (int u : to[v]) {
      now += DP[u];
    }
    now /= to[v].size();
    now += 1;
    DP[v] = now;
  }
  double ans = DP[0];
  for (int nv = 0; nv < N -1; nv++) {
    if (to[nv].size() == 1) continue;
    int nu = to[nv][0];
    for (int u : to[nv]) {
      if (DP[nu] < DP[u]) nu = u;
    }

    double now = [&]{
      vector<double> DP(N);
      for (int v = N - 2; v >= 0; --v) {
        double now = 0;
        for (int u : to[v]) {
          if (v == nv && u == nu) continue;
          now += DP[u];
        }
        int deg = to[v].size();
        if (v == nv) deg--;
        now /= deg;
        now += 1;
        DP[v] = now;
      }
      return DP[0];
    }();
    ans = min(ans, now);
  }

  printf("%.10f\n", ans);
  return 0;
}