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 <iostream>
#include <vector>
#include <cfloat>
#include <iomanip>

using namespace std;
using P = pair<int, int>;

vector<int> table[605];
double dp[605];
double pb[605];

signed main() {
    int N, M;
    cin >> N >> M;
    vector<P> edges;
    for (int i = 0; i < M; i++) {
        int s, t;
        cin >> s >> t;
        s--; t--;
        table[s].push_back(t);
        edges.push_back(P(s, t));
    }

    dp[N-1] = 0;
    for (int i = N-2; i >= 0; i--) {
        double sum = 0;
        int num = 0;
        for (int next : table[i]) {
            sum += dp[next];
            ++num;
        }
        sum /= num;
        dp[i] = sum + 1;
    }

    pb[0] = 1;
    for (int i = 0; i < N; i++) {
        int sz =table[i].size();
        if (sz == 0) continue;
        double cof = 1.0 / sz;

        for (int next : table[i]) {
            pb[next] += pb[i] * cof;
        }
    }

    double ret = dp[0];
    for (P edge : edges) {
        int i = edge.first;
        int j = edge.second;
        int n = table[i].size();
        if (n == 1) continue;
        double val = dp[0];
        val -= pb[i] * dp[i];
        val += pb[i] * ((n * dp[i] - n - dp[j]) / (n-1) +1);
        ret = min(ret, val);
    }

    cout << setprecision(10) << fixed << ret << endl;
    return 0;
}