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;

using ll = long long;

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int N, M;
  cin >> N >> M;
  vector<vector<int>> to(N, vector<int>());
  for (int i = 0; i < M; ++i) {
    int s, t;
    cin >> s >> t;
    --s, --t;
    to[s].push_back(t);
  }
  vector<double> dp(N, 0);
  for (int i = N - 2; i >= 0; --i) {
    for (int j : to[i]) {
      dp[i] += dp[j];
    }
    dp[i] /= to[i].size();
    ++dp[i];
  }
  double ans = dp[0];
  for (int i = 0; i < N - 1; ++i) {
    if (to[i].size() == 1) continue;
    int taken = -1;
    for (int j : to[i]) if (taken == -1 || dp[taken] < dp[j]) taken = j;
    double now = [&]() {
      dp[N - 1] = 0;
      for (int u = N - 2; u >= 0; --u) {
        dp[u] = 0;
        for (int v : to[u]) {
          if (u == i && v == taken) continue;
          dp[u] += dp[v];
        }
        double deg = to[u].size();
        if (u == i) --deg;
        dp[u] /= deg;
        ++dp[u];
      }
      return dp[0];
    }();
    ans = min(ans, now);
  }
  printf("%.10f\n", ans);
}