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>
int main()
{
  long n ; // max n = 200000
  scanf("%ld",&n) ;

  long a[200000] ;
  long i ;

  for( i = 0 ; i < n ; i++){
    scanf("%ld",&a[i]) ;
  }

  long cnt, p ;
  cnt = 0 ;
  long j = 0 ;
  for( p = 1 ; p <= n ; p++ ){
    for( i = j ; i < n ; i++ ){
      if( a[i] == p ){
        cnt++ ;
        j = i+1 ;
        break ;
      }else if(i == n-1 && a[i] != p ){
        cnt = n+1 ;
        break ;
      }
    }
    if(cnt == n+1 ){
      break ;
    }
  }

  printf("%ld\n", n-cnt) ;

  return 0 ;
}