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 <cstdio>

using namespace std;

int main(){
    int N, M;
    while(cin >> N >> M){
        vector<vector<int>> g(N);
        for (int i = 0; i < M;i++){
            int s, t;
            cin >> s >> t;
            g[s-1].push_back(t-1);
        }
        double res = 1e10;
        for (int i = N - 2; i >= 0;i--){
            vector<double> dp(N, 1e10);
            dp[N - 1] = 0.0;
            for (int j = N - 2; j >= 0;j--){
                double sum = 0.0;
                double mx = 0.0;
                for (auto t : g[j]){
                    sum += dp[t];
                    mx = max(mx, dp[t]);
                }
                dp[j] = 1.0 + sum / g[j].size();
                if(i == j && g[j].size() >= 2){
                    dp[j] = min(dp[j], 1.0 + (sum - mx) / (g[j].size() - 1));
                }
            }
            res = min(res, dp[0]);
        }
        printf("%.8lf\n", res);
    }
}