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: Accepted)

#include <bits/stdc++.h>
//#include <boost/multiprecision/cpp_int.hpp>
//namespace mp = boost::multiprecision;

using namespace std;

const double PI = 3.14159265358979323846;
typedef long long ll;
const double EPS = 1e-9;
#define rep(i, n) for (int i = 0; i < (n); ++i)
//#define rep(i, n) for (ll i = 0; i < (n); ++i)
typedef pair<ll, ll> P;
const ll INF = 10e17;
#define cmin(x, y) x = min(x, y)
#define cmax(x, y) x = max(x, y)
#define ret() return 0;

double equal(double a, double b) {
    return fabs(a - b) < DBL_EPSILON;
}

std::istream &operator>>(std::istream &in, set<int> &o) {
    int a;
    in >> a;
    o.insert(a);
    return in;
}

std::istream &operator>>(std::istream &in, queue<int> &o) {
    ll a;
    in >> a;
    o.push(a);
    return in;
}

bool contain(set<int> &s, int a) { return s.find(a) != s.end(); }

//ofstream outfile("log.txt");
//outfile << setw(6) << setfill('0') << prefecture << setw(6) << setfill('0') << rank << endl;
// std::cout << std::bitset<8>(9);
//const ll mod = 1e10;

typedef priority_queue<ll, vector<ll>, greater<ll> > PQ_ASK;

vector<vector<int>> get_graph(int n, vector<P> &edges) {
    vector<vector<int>> graph(n);
    for (P &p:edges) graph[p.first].push_back(p.second);
    return graph;
}

vector<double> solve(int n, vector<P> &edges) {
    auto g = get_graph(n, edges);

    vector<double> dp(n, 0);
    for (int i = n - 2; i >= 0; i--) {
        double now = 0;
        for (int to : g[i]) {
            now += dp[to];
        }
        now /= g[i].size();
        now += 1;
        dp[i] = now;
    }
    return dp;
}

int get_max_e(vector<int> &tos, vector<double> &dp) {
    if (tos.size() == 1) return -1;
    double ma = 0;
    for (int to : tos) cmax(ma, dp[to]);
    for (int to : tos) if (dp[to] == ma) return to;
    __throw_runtime_error("konaide");
}

vector<P> filter_edge(vector<P> edges, P p) {
    auto it = find(edges.begin(), edges.end(), p);
    assert(it != edges.end());
    edges.erase(it);
    return edges;
}

int main() {
    int n, m;
    cin >> n >> m;

    vector<P> edges(m);
    for (P &p:edges) cin >> p.first >> p.second, p.first--, p.second--;

    vector<vector<int>> graph = get_graph(n, edges);

    vector<double> plain = solve(n, edges);

    double ans = plain[0];

    rep(i, n - 2) {
        int max_e = get_max_e(graph[i], plain);
        if (max_e == -1) continue;
        vector<P> edge_ignore = filter_edge(edges, P(i, max_e));
        assert(edge_ignore.size() + 1 == edges.size());
        double now = solve(n, edge_ignore)[0];
        cmin(ans, now);
    }
    cout << setprecision(20) << ans << endl;

}