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 <iomanip>
#include <fstream>
#include <string>
#include <array>
#include <vector>
#include <deque>
#include <list>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <stack>
#include <queue>
#include <bitset>
#include <tuple>
#include <cmath>
#include <complex>
#include <algorithm>
#include <utility>
#include <regex>
#include <cstdint>
#include <numeric>
#include <functional>
#include <cassert>

using namespace std;
using ll = long long;
using ull = unsigned long long;

constexpr ll INF = 1'000'000'000'000'000'000;
namespace utils{
#define ALL(x) begin(x), end(x)
#define RALL(x) rbegin(x), rend(x)
    template<class T, class Compare> using p_queue = priority_queue<T, vector<T>, Compare>;
    template<class T> using min_queue = p_queue<T, greater<T>>;
    template<class T> using max_queue = p_queue<T, less<T>>;

    template<class T> bool up_min(T& X, const T& A){ if(X > A) {X = A; return true;} return false; }
    template<class T> bool up_max(T& X, const T& A){ if(X < A) {X = A; return true;} return false; }

    ll operator"" _64(unsigned long long x){ return ll(x); }
    ull operator"" _64u(unsigned long long x){ return ull(x); }

    template<class T>
    vector<T> make_mdvector(int n, T t){
        return vector<T>(n, t);
    }
    template<class ...Ts>
    auto make_mdvector(int n, Ts ... ts){
        return vector<decltype(make_mdvector(ts...))>(n, make_mdvector(ts...));
    }

    inline ll bit(int x){
        return 1ll << x;
    }

    template<class T>
    istream& operator>>(istream& is, vector<T>& v){
        for (auto &&x : v) { is >> x; } return is;
    }

    template<class T>
    ostream& operator<<(ostream& os, const vector<T>& v){
        for (auto &&x : v) { os << x << ' '; } cout << endl; return os ;
    }
}
using namespace utils;

class solver{
    istream& is;
    ostream& os;

public:
    solver(istream& I, ostream& O) :is(I), os(O) {}
    ll N, M;
    vector<vector<int>> in, out;
    void run();

};

int main(int argc, char *argv[]) {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout << setprecision(16) << scientific;
#ifdef XELMH
    string test_cases = "test_F.txt";
    cerr << test_cases << " -->" << endl;
    auto fs = fstream(test_cases, fstream::in);
    int loop = 0;
    while(fs) {
        loop++;
        cout << '#' << loop << "#------\n";
        solver(fs, cout).run();
    }
    if(loop <= 1) {
        cout << "===" << endl;
        while(cin) solver(cin, cout).run();
    }
#else
    solver(cin, cout).run();
#endif
    return 0;
}


void solver::run(){
    is >> N >> M;
    if(!is) return;
    out.resize(N);
    for (int ii = 0; ii < M; ++ii) {
        int s, t;
        is >> s >> t;
        s--; t--;
        out[s].push_back(t);
    } // end ii

    double ans = INF;
    for (int i = 0; i < N; ++i) {
        auto dp = make_mdvector(N,  1e+9);
        dp[N-1] = 0; dp[N-1] = 0;
        for (int v = N - 2; v >= 0; --v) {
            vector<double> es0;
            double mxe = 0;
            for (auto &&u : out[v]) {
                es0.push_back(dp[u]+1);
                up_max(mxe, dp[u]+1);
            } // end u
            if(es0.empty()) continue;
            dp[v] = 0;
            for (int t = 0; t < es0.size(); ++t) {
                dp[v] += (es0[t]);
            } // end t
            if(i == v and es0.size() > 1){
                dp[v] -= mxe;
                dp[v] /= es0.size() - 1;
            }
            else dp[v] /= es0.size();
        } // end v
        up_min(ans, dp[0]);
    } // end i

    os << ans << endl;
}