Problem:
Score : 600 points

Problem Statement

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}.

Sample Input 1

4 6
1 4
2 3
1 3
1 2
3 4
2 4

Sample Output 1

1.5000000000

If Aoki blocks the passage from Room 1 to Room 2, Takahashi will go along the path 1 → 3 → 4 with probability \frac{1}{2} and 1 → 4 with probability \frac{1}{2}. E = 1.5 here, and this is the minimum possible value of E.

Sample Input 2

3 2
1 2
2 3

Sample Output 2

2.0000000000

Blocking any one passage makes Takahashi unable to reach Room N, so Aoki cannot block a passage.

Sample Input 3

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

Sample Output 3

3.0133333333

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.Scanner;

public class Main {
  public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);

    int N = scanner.nextInt();
    int M = scanner.nextInt();
    List<Integer>[] exits = new List[N];
    for (int i = 0; i < N; i++) exits[i] = new ArrayList<>();

    int[][] edges = new int[M][2];
    for (int i = 0; i < M; i++) {
      int s = scanner.nextInt() - 1;
      int t = scanner.nextInt() - 1;
      exits[s].add(t);
      edges[i][0] = s;
      edges[i][1] = t;
    }

    double[] e = new double[N];
    for (int j = N - 2; j >= 0; j--) {
      for (int k : exits[j]) e[j] += e[k] + 1;
      e[j] /= exits[j].size();
    }
    double min = e[0];

    for (int i = 0; i < M; i++) {
      if (exits[edges[i][0]].size() <= 1) continue;
      Arrays.fill(e, 0.d);
      exits[edges[i][0]].remove(Integer.valueOf(edges[i][1]));

      for (int j = N - 2; j >= 0; j--) {
        for (int k : exits[j]) e[j] += e[k] + 1;
        e[j] /= exits[j].size();
      }
      if (e[0] < Double.POSITIVE_INFINITY) min = Math.min(min, e[0]);

      exits[edges[i][0]].add(edges[i][1]);
    }
    System.out.println(min);
  }
}