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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

public class Program
{
    static void Main()
    {
        string line = Console.ReadLine().Trim();

        string[] lineRecord = Console.ReadLine().Trim().Split(' ');
        var list = new List<int>();
        foreach (var item in lineRecord)
        {
            list.Add(int.Parse(item));
        }

        var answer = CalcBreak(int.Parse(line), list);
        Console.WriteLine(answer);
    }

    private static int CalcBreak(int N, List<int> rengaNO)
    {
        int count = 0;
        if(!rengaNO.Contains(1)) return -1;

        foreach (var item in rengaNO)
        {
            if (item == count + 1) count++;
        }

        return rengaNO.Count() - count;
    }
}