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 Scala (2.11.7) (judge verdict: Time Limit Exceeded)

import java.io.{BufferedReader, InputStreamReader}

import scala.collection.mutable
import scala.collection.mutable.ListBuffer

class Room(val id: Int) {
  val nextRooms: ListBuffer[Room] = new ListBuffer[Room]
  var memo: mutable.HashMap[Int, Double] = new mutable.HashMap()

  def addNext(next: Room): Unit = {
    nextRooms += next
  }

  def expectation(aoki: Int): Double = {
    memo.getOrElseUpdate(aoki,
      nextRooms.size match {
        case 0 => 0.0
        case 1 => nextRooms.head.expectation(aoki) + 1.0
        case size =>
          if (id == aoki) {
            nextRooms.foldLeft((0.0,0.0)) {
              case ((sum, max), room) => {
                val e = room.expectation(aoki)
                (sum + e, Math.max(e, max))
              }
            } match {
              case (sum, max) => (sum - max) / (size - 1) + 1.0
            }
          } else {
            nextRooms.foldLeft(0.0) {
              case (sum, room) => sum + room.expectation(aoki)
            } / size + 1.0
          }
      }
    )
  }

  override def toString: String = {
    "[" + id + " -> (" + nextRooms.map(_.id).sorted.mkString(",") + ")]"
  }
}


object Main {
  def main(args: Array[String]): Unit = {
    val in = new BufferedReader(new InputStreamReader(System.in))
    val Array(n, m) = in.readLine().split("\\s+").map(_.toInt)

    val rooms = (1 to n).map(new Room(_)).toArray

    (1 to m).foreach { _ =>
      in.readLine().split("\\s+").map(_.toInt) match {
        case Array(s, t) => rooms(s - 1).addNext(rooms(t - 1))
      }
    }

    //    System.err.println("------")

    val minE = (1 until n).foldLeft(Double.MaxValue) {
      case (e, aoki) => Math.min(rooms.head.expectation(aoki), e)
    }

    //    rooms.foreach { room =>
    //      System.err.println(room)
    //    }

    System.out.println(f"$minE%2.10f")
  }
}