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: Wrong Answer)

/* ---------- STL Libraries ---------- */
// IO library
#include <cstdio>

#include <fstream>
#include <iomanip>
#include <ios>
#include <iostream>

// algorithm library
#include <algorithm>

#include <cmath>
#include <numeric>
#include <random>
#include <cstring>

// container library
#include <array>
#include <bitset>
#include <deque>
#include <map>
#include <unordered_map>
#include <queue>
#include <set>
#include <string>
#include <tuple>
#include <vector>
#include <stack>

// type
#include <cstdint>

/* ---------- Namespace ---------- */
using namespace std;

/* ---------- Type ---------- */
using ll = long long;
#define int ll
#define P pair<ll, ll>

/* ---------- Constants  */
const double PI = 3.141592653589793238462643383279;
const ll MOD = 1e9 + 7;
const int INF = 1LL << 55;

struct TopologicalSort {
private:
    int V;
    int E;
    vector<vector<int>> table;
public:
    TopologicalSort(int v, int e) {
        V = v;
        E = e;
        table.resize(v);
    }

    void add_edge(int from, int to) {
        table[from].push_back(to);
    }

    // if the size of vector is less than V, this graph is cyclic
    vector<int> sort() {
        vector<int> sorted_nodes;
        stack<int> st;

        vector<int> indegree(V, 0);
        for (int i = 0; i < V; i++) {
            for (int to : table[i]) {
                indegree[to]++;
            }
        }

        for (int i = 0; i < V; i++) {
            if (indegree[i] == 0) st.push(i);
        }

        while (st.size() > 0) {
            int popped = st.top();
            st.pop();

            sorted_nodes.push_back(popped);

            for (int to : table[popped]) {
                indegree[to]--;
                if (indegree[to] == 0) st.push(to);
            }
        }

        return move(sorted_nodes);
    }
};

struct edge {
    int from;
    int to;
};

vector<int> froms[605];

signed main() {
    int N, M;
    cin >> N >> M;

    vector<edge> edges;
    TopologicalSort ts(N, M);

    for (int j = 0; j < M; j++) {
        int s, t;
        cin >> s >> t;
        s--; t--;
        ts.add_edge(s, t);
        edges.push_back({s, t});
        froms[t].push_back(s);
    }

    auto sorted_nodes = ts.sort();

    edges.push_back({INF, INF});

    double ret = 1e18;
    for (auto e : edges) {
        int s = e.from;
        int t = e.to;

        vector<double> dp(N, INF);
        dp[0] = 0;
        for (int node : sorted_nodes) {
            int num = 0;
            int sum = 0;

            for (int from : froms[node]) {
                if (from == s && node == t) continue;
                if (dp[from] >= INF) continue;
                ++num;
                sum += dp[from] + 1.0;
            }

            if (num > 0 && dp[node] == INF) dp[node] = 0;
            if (num > 0) dp[node] += 1.0 * sum / num;
        }

        ret = min(ret, dp[N-1]);
    }

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