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 <iostream>
#include <vector>
#include <algorithm>
#include <utility>
#include <queue>
#include <set>
#include <stdio.h>
#include <string>
using namespace std;
#define ll long long
#define mod 1000000007;

ll gcd(ll a, ll b){
	return (b > 0)? gcd(b, a % b) : a ;
}

ll lcm(int a, ll b){
	return a / gcd(a, b) * b;
}

int main() {
    int n,a[300000],nowmax,count=1;
    vector<int> b;
    cin >> n;
    for(int i=0;i<n;i++)cin >> a[i];
    nowmax = a[n-1];
    b.push_back(nowmax);
    for(int i = n-2;i >=0;i--){
        if(a[i] < nowmax){
            nowmax = a[i];
            count++;
            b.push_back(nowmax);
        }
    }
    reverse(b.begin(), b.end());
    for(int i=0;i<count;i++){
        if(b[i] != i+1){
            cout << -1 << endl;
            return 0;
        }
    }
    cout << n -count << endl;
    return 0;
}