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)

#include <iostream>
#include <vector>
#include <algorithm>
#include <utility>
#include <queue>
#include <string>
#include <set>
#include <bitset>
#include <unordered_map>
#include <stack>

using namespace std;
using ll = long long;
using Pii = pair<int, int>;
using Pll = pair<ll, ll>;
template <class T>
using Vvec = vector<vector<T> >;

// Karn's algorithm
void Topological_sort(Vvec<int> &to, std::vector<int> &ans)
{
    const int V = to.size();
    // std::vector<int> ans; // 結果
    std::stack<int> st; // 入次数0の頂点
    ans.clear();

    // 入次数のカウント
    std::vector<int> in_deg(V, 0);
    for(auto& u_to: to){
        for(int& v: u_to) in_deg[v]++;
    }

    // 入次数0の頂点追加
    for(int i=0; i<V; i++){
        if(in_deg[i] == 0) st.push(i);
    }

    // 更新
    while(!st.empty()){
        int u = st.top(); st.pop();
        ans.push_back(u);
        for(int &v: to[u]){
            in_deg[v]--;
            if(in_deg[v] == 0) st.push(v);
        }
    }
}

int main()
{
    int N, M;
    cin >> N >> M;
    Vvec<int> graph(N);
    for(int i=0; i<M; i++){
        int u, v;
        cin >> u >> v; u--; v--;
        graph[u].push_back(v);
    }

    // vector<int> ans;
    // Topological_sort(graph, ans);

    Vvec<double> dp(N, vector<double>(2, 0.0));
    for(int i=N-2; i>=0; i--){
        double m1=0.0, m2=0.0;
        for(int& to: graph[i]){
            dp[i][0] += dp[to][0];
            m1 = max(dp[to][0], m1);
            m2 = max(dp[to][0] - dp[to][1], m2);
        }
        dp[i][1] = (dp[i][0] - m2) / graph[i].size();
        if(graph[i].size() > 1)
            dp[i][1] = min((dp[i][0] - m1) / (graph[i].size()-1), dp[i][1]);
        dp[i][0] /= graph[i].size();
        dp[i][0]++;
        dp[i][1]++;
    }

    printf("%.8f\n", dp[0][1]);
}