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;
  scanf("%d %d", &n, &m);
  vector<vector<int>> e(n);
  for (int i = 0; i < m; i++) {
    int s, t;
    scanf("%d %d", &s, &t);
    e[s - 1].emplace_back(t - 1);
  }

  double ans = n - 1;
  for (int i = 0; i + 1 < n; i++) {
    if (e[i].size() < 2) continue;

    vector<double> f(n);
    f[n - 1] = 0;
    for (int j = n - 2; j >= 0; j--) {
      double s = 0, maxp = 0; 
      for (int k = 0; k < e[j].size(); k++) {
        double v = f[e[j][k]];
        s += v;
        maxp = max(v, maxp);
      }
      if (j == i)
        f[j] = (s - maxp) / (e[j].size() - 1) + 1;
      else
        f[j] = s / e[j].size() + 1;
    }

    ans = min(f[0], ans);
  }
  printf("%lf\n", ans);

  return 0;
}