Problem: 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}.

Examples

Input

4 6
1 4
2 3
1 3
1 2
3 4
2 4


Output

1.5000000000


Input

3 2
1 2
2 3


Output

2.0000000000


Input

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


Output

3.0133333333

Solution (in CPP):

Solution:
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int main()
{
    int N, M;
    cin>>N>>M;
    vector<vector<int>> E(N);
    for (int i=0; i<M; i++)
    {
        int s, t;
        cin>>s>>t;
        E[s-1].push_back(t-1);
    }

    vector<double> T(N);
    T[N-1] = 0;
    for (int i=N-2; i>=0; i--)
    {
        for (int e: E[i])
            T[i] += T[e]+1;
        T[i] /= E[i].size();
    }

    vector<double> P(N);
    P[0] = 1.0;
    for (int i=0; i<N; i++)
        for (int e: E[i])
            P[e] += P[i]/E[i].size();

    double ans = T[0];
    for (int i=0; i<N; i++)
        if (E[i].size()>1)
        {
            double s = 0;
            double m = 0;
            for (int e: E[i])
            {
                s += T[e]+1;
                m = max(m, T[e]+1);
            }
            double t = T[i] - (s-m)/(E[i].size()-1);
            ans = min(ans, T[0]-t*P[i]);
        }

    printf("%.10f\n", ans);
}