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 <bits/stdc++.h>

using namespace std;

#define REP(i,n) for(int i = 0; i < (int)(n); i++)

double dp[800];

int main(){
  int n, m;
  cin >> n >> m;
  vector<vector<int> > a(n);
  REP(i, m){
    int p, q;
    cin >> p >> q;
    --p, --q;
    a[p].push_back(q);
  }
  double mini = 1e40;
  REP(v, n) if(v == n - 1 || a[v].size() > 1){
    for(int i = n - 2; i >= 0; --i){
      double x = a[i].size();
      double m = 0.;
      double s = 0.;
      REP(j, a[i].size()){
        int d = a[i][j];
        s += dp[d];
        if(m < dp[d]){
          m = dp[d];
        }
      }
      if(i == v){
        s -= m;
        x -= 1.;
      }
      dp[i] = s / x + 1.;
    }
    mini = min(mini, dp[0]);
  }
  printf("%.10f\n", mini);
  return 0;
}