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 <string.h>
#include <math.h>
#include <stdlib.h>

int main(void){
  int n;
  scanf("%d", &n);
  int i, j, k, l;
  int *a = (int *)malloc(n * sizeof(int));
  int ans = -1;
  int max = 0, min;
  for(i=0; i<n; i++){
    scanf("%d", a+i);
    if(max < a[i])
      max = a[i];
    if(i==0)
      min = a[0];
    else if(a[i] < min)
      min = a[i];
  }

  if(min > 1){
    printf("-1\n");
    return 0;
  }
  
  int count = 0;

  /*
  for(i=0; i<n; i++)
    for(j=i+1; j<n; j++)
      if(a[j] != 0){
	
	if(a[i] == a[j]){
	  a[j] = 0;
	}
	
	if(a[j] == l){
	  for(k=l-1; k<j; k++)
	    a[k] = 0;
	}
      }
  */
  j = 0;
  k = 0;
  for(i=1; i<max; i++){
    j = j+1;
    while(a[j] != i && j < max)
      j++;
    while(k<j){
      a[k] = 0;
      k++;
    }
    /*
    printf("i = %d, j = %d, k = %d\n", i, j, k);

    for(int m=0; m<n; m++)
      printf("%d ", a[m]);
    puts("");
    */
    if(j == max)
      break;
    j++;
    k++;
  }
  
  for(i=0; i<n; i++)
    if(a[i] == 0)
      count++;

  printf("%d\n", count);
  
  return 0;
  
}