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 <stdio.h>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long ll;

int main(void) {
  ll i, j, k, n, m, s, t;
  scanf("%lld%lld", &n, &m);
  double p[n], ans, e[n];
  vector<ll> vec[n], rev[n];
  ll cnt[n];
  for(i = 0; i < n; ++i) cnt[i] = 0;
  for(i = 0; i < m; ++i) {
    scanf("%lld%lld", &s, &t);
    vec[--s].push_back(--t);
  }
  for(i = 0; i < n; ++i) p[i] = e[i] = 0.0;
  p[0] = 1.0, e[0] = -1.0;
  for(i = 0; i < n - 1; ++i) for(j = 0; j < vec[i].size(); ++j) p[vec[i][j]] += p[i] / vec[i].size();
  for(i = 0; i < n - 1; ++i) {
    e[i] += 1.0;
    for(j = 0; j < vec[i].size(); ++j) e[vec[i][j]] += e[i] * p[i] / p[vec[i][j]] / vec[i].size();
  }
  ans = e[n - 1] + 1;
  for(i = 0; i < n - 1; ++i) sort(vec[i].begin(), vec[i].end());
  for(k = 0; k < n - 1; ++k) if(vec[k].size() != 1) {
    for(i = 0; i < n; ++i) p[i] = e[i] = 0.0;
    p[0] = 1.0, e[0] = -1.0;
    for(i = 0; i < n - 1; ++i) for(j = 0; j < vec[i].size(); ++j) if(j || i != k) p[vec[i][j]] += p[i] / (i == k ? vec[i].size() - 1 : vec[i].size());
    for(i = 0; i < n - 1; ++i) {
      e[i] += 1.0;
      for(j = 0; j < vec[i].size(); ++j) if(j || i != k) e[vec[i][j]] += e[i] * p[i] / p[vec[i][j]] / (i == k ? vec[i].size() - 1 : vec[i].size());
    }
    if(ans > e[n - 1] + 1) ans = e[n - 1] + 1;
  }
  printf("%lf", ans);
  return 0;
}