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: Wrong Answer)

using System;
using static System.Console;
using static System.Math;
 
public class Hello{
    public static void Main(){
        int max = rint();
        int[] read = inta();
        int breaking = -1;
        for(int i=0;i<max;i++){
            if(breaking == -1 && read[i] == 1){
                breaking = 1;
            }else if(read[i] == breaking + 1){
                breaking++;
            }
        }
        WriteLine(max - breaking);
    
    
    
    
    
    
    }
    //各種読取
    static string rstr(){
        return ReadLine();
    }
    static int rint(){
        return int.Parse(ReadLine());
    }
    static long rlong(){
        return long.Parse(ReadLine());
    }
    static string[] stra(){
        return ReadLine().Split(' ');
    }
    static int[] inta(){
        string[] read_str_array = ReadLine().Split(' ');
        int[] return_int_array = new int[read_str_array.Length];
        for(int countup_i=0;countup_i<read_str_array.Length;countup_i++){
            return_int_array[countup_i] = int.Parse(read_str_array[countup_i]);
        }
        return return_int_array;
    }
    static int[,] inta2(int num_array,int in_array){
        int[,] int_array2 = new int[num_array,in_array];
        for(int i=0;i<num_array;i++){
            int[] temp_array = inta();
            for(int j=0;j<in_array;j++){
                int_array2[i,j] = temp_array[j];
            }
        }
        return int_array2;
    }
    // A/B小数点以下切上
    static int divideup(int divided,int divid){
        int temp_divide = divided/divid;
        if(divided % divid > 0){
            temp_divide++;
        }
        return temp_divide;
    }
}