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;


int main()
{
    ios_base::sync_with_stdio(false);
    long long int a,b,c,i,j,k,v,n;
    cin>>n;
    vector <long long int> v1;
    vector <long long int>::iterator it;
    for(i=0;i<n;i++)
    {
        cin>>v;
        v1.push_back(v);
    }
    k=1;
    i=0;
    while(i<v1.size())
    {
        if(v1[i]==k)
        {
            i++;
            k++;
        }
        else
        {
            it=v1.begin()+i;
            v1.erase(it);
        }
    }
    a=n-v1.size();
    if(a!=n)
    cout<<a<<"\n";
    else
        cout<<"0\n";
    return 0;
}