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 (Clang 3.8.0) (judge verdict: Accepted)

#include <iostream>
using namespace std;
#include <vector>

int main() {

    int n;
    cin >> n;

    vector<int> vec(n);

    for (int i = 0; i < n; i++) {
        cin >> vec.at(i);
    }

    int pre_n = 1;

    int flg = 0;

//    for(auto i=vec.begin(); )

    for (int i = 0; i < n; i++) {
        if (vec.at(i) == 1) {
            flg += 1;
        }
        if (vec.at(i) == pre_n) {
            pre_n += 1;
//            continue;
        }
//        else if (vec.at(i) != pre_n) {
//            vec.erase(vec.begin() +i);
////            vec.push_back()
//            continue;
//        }
    }

    if (flg) {
        cout << n - pre_n + 1;
    }
    else {
        cout << -1;
    }


// TODO:
// until find 1 ok remove previous num
// count remove num
// then find 2 ... remove previous num
// go to last
// check i = n
// if i = n => cout count
// if i != n => cout -1


    return 0;
}