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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#include <iostream>
using namespace std;
#define _LL long long
#define rep(i, n) for (_LL i = 0; i < (_LL)(n); i++)
#define vecrep(itr, v) for (auto itr = (v).begin(); itr != (v).end(); itr++)

int main()
{
    _LL n, m;
    cin >> n >> m;
    vector<_LL> *r = new vector<_LL>[n];
    double *dp = new double[n];
    double *p = new double[n];
    rep(i, m)
    {
        _LL s, t;
        cin >> s >> t;
        s--; t--;
        r[s].push_back(t);
    }
    rep(j, n)
    {
        _LL k = n - 1 - j;
        dp[k] = 0;
        if( j > 0 ) 
        {
            _LL rnum = r[k].size();
            double sum = 0;
            vecrep(itr, r[k])
            {
                sum += dp[*itr];
            }
            dp[k] = sum / rnum + 1;
        }
    }

    rep(k, n) p[k] = 0;
    p[0] = 1;
    double bestdiff = 0;
    rep(k, n)
    {
        double worst = 0;
        double sum = 0;
        _LL rnum = r[k].size();
        vecrep(itr, r[k])
        {
            worst = max(worst, dp[*itr]);
            sum += dp[*itr];
            p[*itr] += p[k] / rnum;
        }
        if( rnum > 1 )
        {    
            double diff = (sum / rnum - (sum - worst) / (rnum - 1));
            bestdiff = max( p[k] * diff, bestdiff);
        }
    }
    cout << dp[0] - bestdiff << endl;
}