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 <iostream>
#include <vector>
#include <cmath>

int main()
{
    int N, M;
    std::cin >> N >> M;
    std::vector<std::vector<int> > link(N);
    std::vector<std::vector<int> > back(N);
    for (int i=0; i<M; ++i) {
        int s, t;
        std::cin >> s >> t;
        link[s-1].push_back(t-1);
        back[t-1].push_back(s-1);
    }

    std::vector<double> v(N);
    v[N-1] = 0;
    for (int i=N-2; i>=0; --i) {
        double sum = 0;
        for (int l=0; l<link[i].size(); ++l) {
            int t = link[i][l];
            sum += v[t] + 1;
        }
        v[i] = sum / link[i].size();
    }

    // turn of Aoki
    int link0 = link[0].size();
    if (link0 == 1) {
        std::cout << v[0] << std::endl;
    } else {
        {
        double maxv = 0;
        for (int i=0; i<link0; ++i) {
            maxv = std::max(v[link[0][i]], maxv);
        }
        double newv0 = (v[0] * link0 - maxv - 1) / (link0 - 1);
        printf("%20.20lf\n", newv0);
        std::cout << newv0 << std::endl;
        }
    }
    return 0;
}