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 <algorithm>
#include <iostream>
#include <vector>
using namespace std;

double dfs(const vector<vector<int>> &G, int v, vector<double> &E){
  if(E[v] >= 0) return E[v];
  double ret = 0;
  size_t d = G[v].size();
  for(size_t i = 0; i < d; ++i){
    int v_ = G[v][i];
    double p = dfs(G,v_,E);
    ret += (p+1)/d;
  }
  E[v] = ret;
  return E[v];
}

int main(){
  int N, M;
  cin >> N >> M;
  vector<vector<int>> G(N);
  vector<int> D(N);
  for(int i = 0; i < M; ++i){
    int s, t;
    cin >> s >> t;
    --s,--t;
    G[s].push_back(t);
    ++D[t];
  }
  vector<double> P(N,0);
  P[0] = 1;
  for(int i = 0; i < N; ++i){
    for(auto v : G[i]){
      P[v] += P[i]/G[i].size();
    }
  }
  vector<double> E(N,-1);
  dfs(G,0,E);
  double ans = E[0];
  for(int i = 0; i < N; ++i){
    int d = G[i].size();
    if(d == 1) continue;
    for(auto v : G[i]){
      double t = E[0] - (E[i] - (E[i]*d-(E[v]+1))/(d-1))*P[i];
      //cerr << i << " " << v << " " << t << endl;
      ans = min(ans,t);
    }
  }
  printf("%.12f\n",ans);
}