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;
using pii = pair<int, int>;

int n, m;
vector<int> adj[666];

double cache[666][666];

double dp(int v, int b) {
  if(v == n) return 0;
  if(adj[v].size() == 0) return 1e18;
  if(cache[v][b] != 0) return cache[v][b];

  vector<double> e;
  int d = adj[v].size();
  for(int u : adj[v]) {
    e.push_back(dp(u, b));
  }

  if(v == b) {
    sort(e.begin(), e.end());
    e.pop_back();
    d--;
  }

  double res = 0;
  for(double t : e) res += t;
  res /= d;
  res += 1.0;

  return cache[v][b] = res;
}

int main(int argc, char **argv) {
#ifdef LOCAL
  freopen("in.txt", "r", stdin);
#endif

  ios_base::sync_with_stdio(false);
  cin.tie(0);

  cin >> n >> m;
  for(int i=0; i<m; i++) {
    int x, y; cin >> x >> y;
    adj[x].push_back(y);
  }

  double res = 1e18;
  for(int i=0; i<=n; i++) {
    if(adj[i].size() == 1) continue;
    double t = dp(1, i);
    res = min(res, t);
  }

  cout.precision(9);
  cout << fixed;
  cout << res << endl;

  return 0;
}