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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

const int N = 500009;
const int mod = 1000000007;

#define IOS                               \
	ios_base::sync_with_stdio(false); \
	cin.tie(0);                       \
	cout.tie(0);
#define int long long int
#define endl "\n"
#define sz(a) (int)((a).size())
#define all(a) a.begin(), a.end()

int n;
int a[N];

bool check(int x)
{
	int found = 0;
	for (int i = 0; i < n; i++)
	{
		if (a[i] == (found + 1))
			found++;
	}
	if (found >= x)
		return 1;
	return 0;
}

int32_t main()
{
	IOS;
	cin >> n;
	for (int i = 0; i < n; i++)
	{
		cin >> a[i];
	}
	int l = 1, r = n;
	while (l < r)
	{
		if (check((l + r + 1) / 2))
			l = (l + 1 + r) / 2;
		else
			r = (((l + 1 + r) / 2) - 1);
	}
	cout << (n - l) << endl;
}