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: Accepted)

#include <stdio.h>
#define N 200000
int main() {
	int num,i,count,nowNum;
	int a[N];
	int flag=0;
	scanf("%d",&num);
	for(i=0;i<num;i++) {
		scanf("%d ",&a[i]);
	}
	
	for(i=0;i<num;i++) {
		if(a[i]==1) {
			flag=1;
			break;
		}
	}
	
	if(flag==0) {
		printf("-1");
		return 0;
	}
	count=0;
	nowNum=1;
	for(i=0;i<num;i++) {
		if(a[i]==nowNum) {
			nowNum++;
		} else {
			count++;
		}
	}
	printf("%d",count);
	return 0;
}