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 <algorithm>
#include <cassert>
#include <iomanip>
#include <iostream>
#include <vector>
using namespace std;

namespace IO {
    const int BUFFER_SIZE = 1 << 15;

    char input_buffer[BUFFER_SIZE];
    int input_pos = 0, input_len = 0;

    void _update_input_buffer() {
        input_len = fread(input_buffer, sizeof(char), BUFFER_SIZE, stdin);
        input_pos = 0;

        if (input_len == 0)
            input_buffer[0] = EOF;
    }

    inline char next_char(bool advance = true) {
        if (input_pos >= input_len)
            _update_input_buffer();

        return input_buffer[advance ? input_pos++ : input_pos];
    }

    template<typename T>
    inline void read_int(T &number) {
        bool negative = false;
        number = 0;

        while (!isdigit(next_char(false)))
            if (next_char() == '-')
                negative = true;

        do {
            number = 10 * number + (next_char() - '0');
        } while (isdigit(next_char(false)));

        if (negative)
            number = -number;
    }

    template<typename T, typename... Args>
    inline void read_int(T &number, Args &... args) {
        read_int(number);
        read_int(args...);
    }
}


int N, M;
vector<vector<bool>> adj;
vector<int> out;
vector<double> dp;
vector<double> probability;

double solve_dp() {
    dp.assign(N, 0);

    for (int i = N - 2; i >= 0; i--) {
        double sum = 0;

        for (int j = i + 1; j < N; j++)
            if (adj[i][j])
                sum += dp[j] + 1;

        dp[i] = sum / out[i];
    }

    return dp[0];
}

void solve_probability() {
    probability.assign(N, 0);
    probability[0] = 1;

    for (int i = 0; i < N; i++)
        for (int j = i + 1; j < N; j++)
            if (adj[i][j])
                probability[j] += probability[i] / out[i];
}

int main() {
    cout << fixed << setprecision(12);

    IO::read_int(N, M);
    adj.assign(N, vector<bool>(N, false));
    out.assign(N, 0);

    for (int i = 0; i < M; i++) {
        int s, t;
        IO::read_int(s, t);
        s--; t--;
        adj[s][t] = true;
        out[s]++;
    }

    double base = solve_dp();
    double best = base;
    solve_probability();

    for (int s = 0; s < N; s++) {
        if (out[s] <= 1)
            continue;

        double worst = 0;

        for (int t = s + 1; t < N; t++)
            if (adj[s][t])
                worst = max(worst, dp[t]);

        double save = (worst - (dp[s] - 1)) / (out[s] - 1);
        best = min(best, base - probability[s] * save);
    }

    cout << best << '\n';
}