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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.*;
import java.util.*;
import static java.lang.System.*;
import static java.lang.Math.*;

public class Main
{
	public static void main(String []args)throws IOException
	{
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		int N = Integer.parseInt(br.readLine());
		String []s=br.readLine().split(" ");
		int a[]=new int[N];
		int k=1,c=0;
				
		for(int i=0;i<N;i++)
			a[i]=Integer.parseInt(s[i]);
			
		for(int i=0;i<N;i++)
		{
			if(a[i]==k)
			{
				++k;
				++c;
			}
		}
		
		if(c==0)
			out.println(-1);
		else
			out.println(N-c);
	}
}