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<cstdio>
#include<cmath>
#include<climits>
#include<algorithm>
#include<sstream>
#include<iostream>
#include<string>
#include<queue>
#include<stack>
#include<set>
#include<list>
#include<map>
#include<vector>

using namespace std;


#define rep(i,n) for(int i=0; i<n; i++)
#define rept(i,a,b) for(int i=a; i<b; i++)
#define ll long long
#define MOD 1000000007

#define IMIN INT_MIN;
#define IMAX INT_MAX;


int main(){
    int N;
    cin >> N;
    vector<int> a(N);
    rep(i, N) cin >> a[i];
    bool flag = false;
    rep(i, N) if (a[i] == 1) flag = true;
    if (!flag) cout << -1 << endl;
    else {
        int cnt = 1;
        rep(i, N) {
            if (a[i] == cnt) cnt++;
        }
        cout << N - cnt + 1 << endl;

    }


	return 0;
}