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;
typedef long long ll;
typedef long double ld;

const int N = 606;
ld E[N];
vector<int> ch[N];
int n, m;

int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  cin >> n >> m;
  for (int i=1; i<=m; ++i) {
    int u, v;
    cin >> u >> v;
    ch[u].push_back(v);
  }

  ld ans = 1e30;

  for (int x=1; x<=n; ++x) {
    E[n] = 0;
    for (int u=n-1; u>=1; --u) {
      ld esum = 0;
      for (auto& v : ch[u]) {
        esum += E[v];
      }
      E[u] = esum/ch[u].size();

      if (u == x && ch[u].size() > 1) {
        ld emax = 0;
        for (auto& v : ch[u]) {
          emax = max(emax, E[v]);
        }
        E[u] = min(E[u], (esum-emax)/(ch[u].size()-1));
      }

      E[u] += 1;
    }
    ans = min(ans, E[1]);
    cerr << " x " << x << " E " << E[1] << endl;
  }
  cout << fixed << setprecision(10) << ans << endl;
  return 0;
}