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;
 
int main() {
  long long N;
  cin >> N;
  vector<long long> a(N);
  for(long long i = 0; i < N; i++){
    cin >> a[i];
    }
  int Z = 1;
  for(long long i = 0; i < N; i++){
    if(i + 1 != a [i]){
      Z = 0;
      }
    }
  if(Z == 1){
    cout << 0 <<endl;
    }
  
  else{
  
  long long i = 1;
  long long j = 0;
  long long S = 0;
  long long b = 1;
  while(j != N - 1){
    if(i == a[j]){
      i += 1;
      S += j - b + 1;
      b = j;
    }
    j++;
  }
  if (S  != 0){
  cout << S << endl;
    }
  else{
  cout << -1 << endl;
    }
    }
}