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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>

int	main(void)
{
  	int	N;
  	int	*a;
  	int	num;
  	int i;
  	int j;
  	int	flag;
  	int place;
  
  	num = scanf("%d", &N);
  	if (num != 1)
      	return (0);
  
  	a = (int *)malloc(sizeof(int) * (N + 1));
  	for (i=1; i<=N; i++)
    {
      	num = scanf("%d", &a[i]);
      	if (num != 1)
          	return (0);
    }

    flag = 0;
    for (i=1; i<=N; i++)
    {
        if (a[i] == 1)
        {
            place = i;
            flag = 1;
        }
    }
    if (flag == 0)
    {
        printf("-1");
        return (0);
    }
    
  	for (i=place; i<=N; i++)
    {
        for (j=1; j<=N; j++)
        {
            if (a[i] == j)
            {
                place = j;
                break ;
            }
        }
    }
  	printf("%d", N - j);
  	return (0);
}