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 <algorithm>
#include <vector>
#include <iomanip>

#define ld long double

using namespace std;

signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    int n, m;
    cin >> n >> m;
    vector<vector<int>> gr(n);
    vector<ld> ans(n);
    for (int i = 0, x, y; i < m; i++)
    {
        cin >> x >> y;
        x--, y--;
        gr[x].push_back(y);
        gr[y].push_back(x);
    }

    for (int i = n - 2; i >= 0; i--)
    {
        ld sum = 0;
        for (auto each : gr[i])
            sum += ans[each] + 1;
        sum /= gr[i].size();
        ans[i] = sum;
    }

    ld res = ans[0];

    for (int block = 0; block < n - 1; block++)
    {
        if (gr[block].size() == 1) continue;

        ans.clear();
        ans.resize(n);

        for (int i = n - 2; i >= 0; i--)
        {
            ld sum = 0, biggest = 0;
            for (auto each : gr[i])
            {
                sum += ans[each] + 1;
                biggest = max(biggest, ans[each] + 1);
            }

            if (i == block) sum = (sum - biggest) / (gr[i].size() - 1);
            else sum /= gr[i].size();

            ans[i] = sum;
        }

        res = min(res, ans[0]);
    }

    cout << fixed << setprecision(20);
    cout << res << "\n";

    return 0;
}