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 <iomanip>
using namespace std;
typedef long long ll;

double INF = 1e9;

int main() {
  int n, m; cin >> n >> m; vector<int> v[n]; // tiの方
  double an[n]; fill(an, an+n, INF); an[n-1] = 0;
  for(int i=0; i<m; i++) {
    int a, b; cin >> a >> b; a--; b--;
    v[a].push_back(b);
  }
  double anan = 1e16;
  for(int id=n-2; id>=0; id--){
    fill(an, an+n, INF); an[n-1] = 0;
  for(int i=n-2; i>=0; i--) {
    double sm=0, mx=0;
    int lg = v[i].size();
    for(int x:v[i]) {
      sm += an[x] + 1; mx = max(mx, an[x] + 1);
    }
    
    if (id == i && lg >= 2) {
      double ax = (double)sm / (double)lg;
      double bx = (double)((double)sm - (double)mx) / (double)((double)lg - (double)1);
      an[i] = min(ax, bx);
    } else an[i] = (double)sm / (double)lg;
    //an[i] = (double)sm / (double)lg;
    //cout << i << " " << sm << " " << mx << " " << lg << " " << an[i] <<  endl;
  }
    anan = min(anan, an[0]);
  }
  //for(int i=0; i<n; i++) {
  //  cout << an[i] << endl;
  //}
  cout << fixed << setprecision(7);
  cout << anan << endl;
}