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# (.NET Core 3.1.201) (judge verdict: Wrong Answer)

using System;
using System.IO;
using System.Linq;

class Program
{
    static void Main(string[] args) {
        var writer = new StreamWriter(Console.OpenStandardOutput()) {AutoFlush = false};
        Console.SetOut(writer);
        Console.WriteLine(Solve());
        Console.Out.Flush();
    }

    static object Solve() {
        var N = int.Parse(Console.ReadLine());
        var a = Array.ConvertAll(Console.ReadLine().Split(), int.Parse);
        var aLen = a.Length;
        var b = a.Reverse().Distinct().Reverse().ToArray();
        var bLen = b.Length;
        var count = 0;
        var i = 1;
        for (int j = 0; j < bLen; j++) {
            if (!b.Contains(i)) return -1;
            if (i != b[j]) count++;
            else i++;
        }
        return aLen - bLen + count;
    }
}