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;
#define int long long

int N;
vector<int> A;
vector<set<int>> vs;

signed main() {
  cin >> N;
  A.resize(N);
  for(int i=0; i<N; i++) cin >> A[i];
  for(int i=0; i<N; i++) A[i]--;

  vs.resize(N);
  for(int i=0; i<N; i++) vs[A[i]].insert(i);

  if(vs[0].size()==0)  {
    cout <<"-1" << endl;
    return 0;
  }

  int ans = 0;
  int pos = -1;
  for(int a=0; a<N; a++){
    if(vs[a].size()==0) {
      ans += N-pos-1;
      break;
    }
    for (auto i : vs[a] ) {
      if(pos<i) {
        ans += i-pos-1;
        pos=i;
        break;
      }
    }
  }
  cout << ans << endl;  
}