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>
using namespace std;

int main() {

  ios::sync_with_stdio(false);
  cout.tie(nullptr);
  cin.tie(nullptr);
  cout << fixed << setprecision(10);
  
  int n, m;
  cin >> n >> m;

  vector<vector<int>> g(n + 5);
  for (int i = 0; i < m; ++i) {
    int u, v;
    cin >> u >> v;

    g[u].push_back(v);
  }

  auto calc = [&](vector<double> &d, int x) {
    for (auto to : g[x])
      d[x] += d[to];
    d[x] /= g[x].size();
    d[x]++;
  };

  vector<double> dp(n + 5, 0);
  for (int i = n - 1; i > 0; --i) {
    calc(dp, i);
  }

  double ans = dp[1];
  for (int i = n - 1; i > 0; --i) {
    if (g[i].size() == 1) continue;
    vector<double> cur(n + 5, 0);
    double mx = 0;
    for (auto to : g[i])  mx = max(mx, dp[to]);
    for (int j = n - 1; j > 0; --j) {
      if (j > i)  cur[j] = dp[j];
      else if (j == i)  {
        if (g[j].size() == 1) cur[j] = 0;
        else cur[j] = (dp[j] * g[j].size() - mx - 1) / (g[j].size() - 1);
      }
      else calc(cur, j);
    }
    ans = min(ans, cur[1]);
  }
  cout << ans << '\n';  
}