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: Wrong Answer)

#include <bits/stdc++.h>
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>

using namespace std;

long long MOD = 1e9+7;
int MAX_INT = 1e9;

int main(void){
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);

  int N;
  cin >> N;
  vector<int> a(N);
  for(int i = 0; i < N; i++) {
    cin >> a[i];
  }

  int cnt = 0, place = 0;

  for(int i = 1; i <= N; i++) {
    for(int j = place; j < N; j++) {
      if(i == a[j]) {
        place = j + 1;
        break;
      }
      else {
        cnt++;
      }
    }
  }

  if(cnt > N) cout << -1 << endl;
  else cout << cnt << endl;

  return 0;
}