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<cstdlib>
#include<queue>
#include<set>
#include<vector>
#include<string>
#include<cmath>
#include<algorithm>
#include<stack>
#include<map>
#include<deque>
#include<cstdio>
#include<cstdio>
using namespace std;
#define rep(i,a) for(int i=0;i<a;i++)
#define mp make_pair
#define pb push_back
#define ll __int64
#define __int64 long long
#define P pair<int,int>
#define PP pair<P,int>

int n;
int a[222222];

int main(){
    cin>>n;
    rep(i,n)cin>>a[i];
    int cnt=0,pos=1;

    rep(i,n){
        if(a[i]==pos){
            pos++;
            cnt++;
        }
    }

    if(cnt==0){
        cout<<-1<<endl;
    }
    else{
        cout<<n-cnt<<endl;
    }
    return 0;
}