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++14 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <iostream>
#include <string>
#include <vector>
#include <utility>
#include <algorithm>

using namespace std;

long long euclidean_gcd(long long a, long long b)
{
	if (a < b)
		return euclidean_gcd(b, a);
	long long r;
	while ((r = a % b))
	{
		a = b;
		b = r;
	}
	return b;
}

int main()
{
	int n;
	cin >> n;
	vector<int> a(n, 0);
	for (int i = 0; i < n; i++)
		cin >> a[i];
	int r = 0;
	int last = -1;

	for (int i = 0, m = 1; i < n; i++)
	{
		if (a[i] == m)
		{
			m++;
			last = i;
		}
		else
			r++;
	}

	if (last == -1)
		r = -1;
	else
		r -= (n - last - 1);
	printf("%d", r);

	return 0;
}