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 rep(i, n) for (int i = 0; i < (int)(n);i++)
using ll = long long;
int main(){
    int n,m,j=1;
    int f = 0;
    int k=0;
    vector<int> a,b;
    cin >> n;

    rep(i,n){
        cin >> m;
        a.push_back(m);
        if(a.at(i)==j){
            j++;
            f++;
            k = i;
            b.push_back(k);
        }
    }
    


    if(f==0){
        cout << -1 << endl;
    }else{
        int x=0;

        rep(i,b.size()){
            if(i==0){
                x = b[i];
            }
            x += b[i] - b[i - 1];
        }

        int c = b.back() + 1 - a.at(b.back());
        cout << c << endl;
        
    }

}