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<iostream>
#include<vector>
#include<string>
#define rep(i, start, end) for (int i = (int)start; i < (int)end; ++i)
#define rrep(i, start, end) for (int i = (int)start - 1; i >= end; --i)
#define iter(i, c) for (auto i = (c).begin(); i != (c).end(); ++i)
#define riter(i, c) for (auto i = (c).rbegin(); i != (c).rend(); ++i)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
using namespace std;
using ll = long long;

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N;
    cin >> N;
    vector<int> A(N);
    rep(i, 0, N) cin >> A[i];
    bool is_ok = false;
    rep(i, 0, N) {
        if (A[i] == 1) {
            is_ok = true;
            break;
        }
    }
    if (!is_ok) {
        cout << -1 << endl;
        return 0;
    }
    int now = 1;
    rep(i, 0, N) {
        if (A[i] == now) ++now;
    }
    cout << N - now + 1 << endl;
    return 0;
}