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>
using namespace std;
#define int long long
#define PI 3.141592653589793
#define rep(i, n) for (int i = 0; i < (n); i++)
#define rrep(i, n, k) for (int i = (n); i >= (k); i--);
#define all(x) (x).begin(), (x).end()
template<class T> istream &operator>>(istream&is,vector<T>&v){for(auto &elemnt:v)is>>elemnt;return is;}
template<class T,class U> istream &operator>>(istream&is,pair<T,U>&p){is>>p.first>>p.second;return is;}
template<class T>vector<T> make_vector(size_t a){return vector<T>(a);}
template<class T, class... Ts>auto make_vector(size_t a, Ts... ts){return vector<decltype(make_vector<T>(ts...))>(a, make_vector<T>(ts...));}
int mod = 1e9 + 7;

signed main() {
    int N;
    cin >> N;
    vector<int> a(N);
    cin >> a;
    int ans = 0, count = 1;
    for (int i = 0; i < N; i++) {
        if (a[i] == count) ++count;
        else {
            ++ans;
        }
    }
    if (ans == N) cout << -1 << endl;
    else cout << ans << endl;
    return 0;
}