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: Runtime Error)

// ref. http://kmjp.hatenablog.jp/entry/2019/10/28/0930
// ref. https://www.hamayanhamayan.com/entry/2019/10/27/235840
// ref. https://maspypy.com/atcoder-%E5%8F%82%E5%8A%A0%E6%84%9F%E6%83%B3-2019-10-27abc-144

import java.util.HashMap;
import java.util.LinkedList;
import java.util.List;
import java.util.Scanner;

public class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int N = sc.nextInt();
    int M = sc.nextInt();
    HashMap<Integer, List<Integer>> nodeToNexts = new HashMap<>(), nodesFromPrev = new HashMap<>();
    for (int i = 0; i < M; i++) {
      int from = sc.nextInt() - 1; // 0-indexed.
      int to = sc.nextInt() - 1;
      nodeToNexts.computeIfAbsent(from, k -> new LinkedList<>()).add(to);
      nodesFromPrev.computeIfAbsent(to, k -> new LinkedList<>()).add(from);
    }

    double[] dpExp = new double[N]; // dpExp[v]:= Expectation(v->N)
    for (int i = N - 2; i >= 0; i--) {
      for (int next : nodeToNexts.get(i)) {
        dpExp[i] += dpExp[next];
      }
      dpExp[i] /= nodeToNexts.get(i).size();
      dpExp[i] += 1;
    }
    double[] dpReachProb = new double[N]; // dpReachProb[v]:= probReach(0->v)
    dpReachProb[0] = 1;
    for (int idxNext = 1; idxNext < N; idxNext++) {
      for (int prev : nodesFromPrev.get(idxNext)) {
        double prob = 1 / (double) nodeToNexts.get(prev).size();
        dpReachProb[idxNext] += dpReachProb[prev] * prob;
      }
    }


    double improve = 0;
    for (int i = 0; i < N - 1; i++) {
      List<Integer> nexts = nodeToNexts.get(i);
      int preCutOutNum = nexts.size();
      if (preCutOutNum <= 1) continue;
      double maxExp = 0;
      double nextExps = 0;
      for (Integer next : nexts) {
        maxExp = Math.max(dpExp[next], maxExp);
        nextExps += dpExp[next];
      }
      double expPreCut = dpExp[i];
      double expPostCut = (nextExps - maxExp) / (preCutOutNum  - 1) + 1;
      double tmpImprove = (expPreCut - expPostCut) * dpReachProb[i]; // ?
      if (improve < tmpImprove)
        improve = tmpImprove;
    }

    System.out.println(dpExp[0] - improve);
  }
}