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 <algorithm>
#include <vector>
#include <cstdio>
#define rep(i,n) for(int i=0;i<n;++i)
#define rep1(i,n) for(int i=1;i<=n;++i)
using namespace std;
int main()
{
  int n,m;
  cin >> n >> m;
  vector<vector<int>> edge(n);
  rep(i,m){
    int s,t;
    cin >> s >> t;
    s--;
    t--;
    edge[s].push_back(t);
  }

  vector<double> e(n,0);
  for(int v=n-2;v>=0;--v){
    double x = edge[v].size();
    for(auto w:edge[v]){
      e[v] += e[w]/x;
    }
    e[v] += 1.;
  }

  // rep(i,n){
  //   cout << i << ":" << e[i] << "\n";
  // }

  double res = 1e+15;
  rep(v,n){
    if(edge[v].size()<=1){
      res = min(res,e[0]);
      continue;
    }
    double mx=0,sum=0,x=edge[v].size();
    vector<double> ee=e;
    for(auto w:edge[v]){
      sum += e[w];
      mx = max(mx,e[w]);
    }
    sum -= mx;
    ee[v] = sum/(x-1.)+1.;
    if(v!=0){
      for(int vv=v-1;vv>=0;--vv){
	ee[vv] = 0.;
	double xx = edge[vv].size();
	for(auto ww:edge[vv]){
	  ee[vv] += ee[ww]/xx;
	}
	ee[vv] += 1.;
      }
    }
    res = min(res,ee[0]);
  }

  printf("%.7f\n", res);
  return 0;
}