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 <bits/stdc++.h>
using namespace std;
#define _GLIBCXX_DEBUG
#define pb push_back
#define fs first
#define sc second
#define all(x) (x).begin(), (x).end()
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define rep1(i,n) for(int i=1;i<=(int)(n);i++)
typedef long long ll;
typedef pair<char, char> c_c;
typedef pair<int, int> i_i;
typedef pair<ll, ll> l_l;
const int INF = 1000000007;
const ll LLINF = 1000000000000000007LL;

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    
    int n;
    cin >> n;
    vector<int> a(n);
    for(int i=0; i<n; i++) cin >> a[i];
    int cnt = 0;
    stack<int> stc;
    rep(i,n){
        if(stc.size() == 0 ){
            if(a[i] == 1) stc.push(a[i]);
            //cout << stc.top() << endl;
        }
        else{
            if(a[i] == stc.top()+1){
                stc.push(a[i]);
            }
        }
        //cout << stc.top() << endl;
    }
    if(stc.size() != 0) cout << n-stc.size() << endl;
    else cout << -1 << endl;
    
    return 0;
}