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: Accepted)

object Main extends App {
  val sc = new java.util.Scanner(System.in)
  val n = sc.nextInt
  val a = new Array[Int](n)
  for (i <- 0 until n) {
    a(i) = sc.nextInt()
  }

  var currentNum = 0
  for(num <- a) {
    if(currentNum + 1 == num) currentNum += 1
  }
  println(if(currentNum == 0) -1 else n - currentNum)

// ↓最初の解答
//  if (!a.contains(1)) {
//    println(-1)
//  } else {
//    var breakCount = 0
//    var currentNum = 1
//    for (i <- 0 until n) {
//      if (a(i) == currentNum) {
//        currentNum += 1
//      } else {
//        breakCount += 1
//      }
//    }
//    println(breakCount)
//  }
}