Problem:
Score : 400 points

Problem Statement

We have N bricks arranged in a row from left to right.

The i-th brick from the left (1 \leq i \leq N) has an integer a_i written on it.

Among them, you can break at most N-1 bricks of your choice.

Let us say there are K bricks remaining. Snuke will be satisfied if, for each integer i (1 \leq i \leq K), the i-th of those brick from the left has the integer i written on it.

Find the minimum number of bricks you need to break to satisfy Snuke's desire. If his desire is unsatisfiable, print -1 instead.

Constraints

All values in input are integers.

1 \leq N \leq 200000

1 \leq a_i \leq N

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

Print the minimum number of bricks that need to be broken to satisfy Snuke's desire, or print -1 if his desire is unsatisfiable.

Sample Input 1

3
2 1 2

Sample Output 1

1

If we break the leftmost brick, the remaining bricks have integers 1 and 2 written on them from left to right, in which case Snuke will be satisfied.

Sample Input 2

3
2 2 2

Sample Output 2

-1

In this case, there is no way to break some of the bricks to satisfy Snuke's desire.

Sample Input 3

10
3 1 4 1 5 9 2 6 5 3

Sample Output 3

7

Sample Input 4

1
1

Sample Output 4

0

There may be no need to break the bricks at all.

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

import java.util.Scanner

import scala.annotation.tailrec

object Main extends App {
  val sc = new Scanner(System.in)
  val t = System.currentTimeMillis

  val n  = sc.nextInt
  val av = Vector.fill(n)(sc.nextInt)

  @tailrec
  def loop( vec: Vector[Int], target: Int = 1 ): Int = {
    val rest = vec.dropWhile(_ != target)
    (target, rest.isEmpty) match {
      case (1, true) => -1
      case (_, true) => target
      case _ => loop(rest.tail, target + 1 )
    }
  }

  val target = loop(av)
  val res = if(target == -1) -1 else n - target + 1

  println(res)

}