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 <cstdio>
#include <vector>
#include <algorithm>

const int MAXN = 600;
const double INF = MAXN;
std::vector<int> edges[MAXN];
double e[MAXN];

double calc(int n, int block) {
  e[n - 1] = 0;
  for (int u = n - 2; u >= 0; --u) {
    double sum = 0.0, max = 0.0;
    for (const auto &v : edges[u]) {
      sum += e[v];
      if (e[v] > max) max = e[v];
    }
    double num = (int)edges[u].size();
    if (u == block && num > 1) {
      --num;
      sum -= max;
    }
    e[u] = sum / num + 1.0;
  }
  return e[0];
}

double solve(int n) {
  double res = INF;
  for (int i = 0; i < n - 1; ++i) {
    double cur = calc(n, i);
    if (cur < res) res = cur;
  }
  return res;
}

int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  for (int i = 0, u, v; i < m; ++i) {
    scanf("%d %d", &u, &v);
    edges[u - 1].emplace_back(v - 1);
  }
  printf("%.10lf\n", solve(n));
  return 0;
}