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 9.2.1) (judge verdict: Accepted)

#include <iostream>
#include <iomanip>
#include <algorithm>
#include <bitset>
#include <string>
#include <cmath>
#include <complex>
#include <numeric>
#include <cassert>

#include <vector>
#include <array>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <deque>
#include <utility>

#define PI 3.14159265358979323846
#define int64 long long
#define uint64 unsigned long long
#define coutfix(i) cout << fixed << setprecision(i)		//coutの浮動小数出力の桁数設定
using namespace std;

int main()
{
	int64 nn, one = 0, ans = 0, memo = 1;
	cin >> nn;
	vector<int64> aa(nn);
	for(int64 ii = 0;ii < nn;ii++)
	{
		cin >> aa[ii];
		if(aa[ii] == 1)
		{
			one = 1;
		}
	}
	if(one == 0)
	{
		cout << -1 << endl;
		return 0;
	}
	for(int64 ii = 0;ii < nn;ii++)
	{
		if(memo == aa[ii])
		{
			memo++;
		}
		else
		{
			ans++;
		}
	}
	cout << ans << endl;
	return 0;
}