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 (Clang 3.8.0) (judge verdict: Accepted)

#include <stdio.h>

/* 数字をn個残してそれが1からn迄昇順に並んでいればOK */

int main()
{
	int n;		/* ユーザー入力の配列の要素数 */
	int a[200000];	/* 最大200000の数値が入る配列 */
	int index;	/* 配列aのインデックス */

	scanf("%d", &n);
	for(index = 0; index < n; ++index)
		scanf("%d", &a[index]);

	/* 1を探索しその右に2,3...があるかを探す */
	
	int number;	/* 現在探索中の数値 */
	int counter;	/* 発見できた要素の数 */
	number = 1;
	counter = 0;
	index = 0;
	
	while(1){
      	if(index == n)
          break;
		if(a[index] == number){
			++counter;
			++number;
		}
      	++index;
	}

	int brick;	/* 砕くべきレンガの数 */
	brick = n - counter;

	if(number == 1)
		printf("-1\n");
	else
		printf("%d\n", brick);

	return 0;
}