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>
#define rep(i,n) for (int i = 0; i < (n); i++)
using namespace std;
typedef long long ll;

int main(){
  int n, m; cin >> n >> m;
  int M[180000][2];
  vector<vector<int>> to(n);
  rep(i,m){
    int s, t; cin >> s >> t;
    s--; t--;
    to[s].push_back(t);
    M[i][0] = s; M[i][1] = t;
  }
  double A[605];
  A[n-1] = 0;
  for (int i=n-2; i>=0; i--){
    double E = 0;
    for (int j : to[i]) E += A[j] + 1;
    A[i] = E / to[i].size();
  }

  double ans = A[0];
  rep(i, n-1){
    if (to[i].size() <= 1) continue;
    double nextmax = 0;
    int t;
    for(int j: to[i]){
      if (A[j] > nextmax){
        nextmax = A[j];
        t = j;
      }
    }
    double B[605];
    for (int j=n-1; j>=0; j--){
      if (j > i){
        B[j] = A[j];
        continue;
      }
      int size = to[j].size();
      if (size > 1 && j == i) size --;
      double E = 0;
      for (int k: to[j]){
        if (j == i && k == t) continue;
        E += B[k] + 1;
      }
      B[j] = E / size;
    }
    ans = min(ans, B[0]);
  }
  printf("%.10f\n",ans);
  return 0;
}