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 <stdio.h>
#include <math.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <numeric>
#include <string>
#include <map>
using namespace std;
typedef long long int lli;
#define urept(soeji, start, n) for (int soeji = start; soeji < n; soeji++)
#define drept(soeji, start, n) for (int soeji = start; soeji > n; soeji--)
int main(void)
{
    int N;
    cin >> N;
    vector<int> x;
    int tmp;
    for (int i = 0; i < N; i++)
    {
        cin >> tmp;
        x.push_back(tmp);
    }
    int element = 1;
    int count = 0;
    bool res = false;
    int sm = x[0];
    for (int i = 0; i < N; i++)
    {

        if (x[i] == element + 1)
        {
            count++;
            element++;
        }
    }
    int *cpa;
    cpa = new int[N];
    for (int i = 0; i < N; i++)
    {
        cpa[i] = x[N - 1 - i];
    }
    int ck = 0;
    for (int i = 1; i < N; i++)
    {
        if (cpa[i - 1] + 1 == cpa[i])
        {
            ck++;
        }
    }
    if (is_sorted(cpa, cpa + N) && ck == N - 1)
    {
        cout << N - 1 << endl;
        return 0;
    }
    bool f = false;
    for (int i = 0; i < N; i++)
    {
        if (x[i] == 1)
        {
            f = true;
        }
    }
    if (f == true)
    {
        cout << N - count + 1;
    }
    else
    {
        cout << -1 << endl;
    }
    return 0;
}