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<iomanip>
#include<vector>
#include<algorithm>
#include<cstdlib>
template<typename T>
void fin(T const& t){ std::cout << t << std::endl; exit(0); }

int const MAXN = 600;
double const INF = 1e30;

std::vector<int> G[MAXN];
double ex[MAXN];
int h;

double calc(int v) {
  if(ex[v] < INF) return ex[v];
  double res = 1;
  double num = 0, mx = 0, sum = 0;
  for(auto c: G[v]) {
    ++num;
    double e = calc(c);
    sum += e;
    mx = std::max(mx, e);
  }
  if(v == h) {
    sum -= mx;
    --num;
  }
  res += sum/num;
  return ex[v] = res;
}

int main() {
  std::cout << std::setprecision(10) << std::fixed;
  int N, M; std::cin >> N >> M;
  for(int i = 0; i < M; ++i) {
    int s, t; std::cin >> s >> t; --s; --t;
    G[s].push_back(t);
  }
  h = -1;
  std::fill_n(ex, N, INF);
  ex[N-1] = 0e0;
  double ans = calc(0);
  for(h = 0; h < N-1; ++h) if(G[h].size() > 1) {
    std::fill_n(ex, N, INF);
    ex[N-1] = 0e0;
    ans = std::min(ans, calc(0));
  }
  fin(ans);
  return 0;
}