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: Accepted)

#include <bits/stdc++.h>
#define ALL(a)  (a).begin(),(a).end()
#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
typedef pair<long long, long long> Pll;
typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef vector<long long> vll;
typedef vector<vector<long long>> vvll;
template <typename T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template <typename T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
const long long INF = 1LL << 60;
const int INT_INF = 1LL << 30;

int main(){
  int N;
  cin >> N;
  vi a(N);
  for(int i = 0; i < N; i++){
    cin >> a.at(i);
  }
  int ans = 0;
  int num = 1;
  for(int i = 0; i < N; i++){
    if(a.at(i) != num) ans++;
    else num++;
  }
  if(ans == N) cout << -1 << endl;
  else cout << ans << endl;
}