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)

#ifndef __cplusplus
    cat << EOF
#endif
#include <bits/stdc++.h>
using namespace std;

#define rep(...) overload_rep(__VA_ARGS__, rep_1, rep_0)(__VA_ARGS__)
#define rep_0(i, end) for(int i = 0; i < (end); ++i)
#define rep_1(i, fr, ba) for(int i = (fr); i <= (ba); ++i)
#define overload_rep(arg0, arg1, arg2, name, ...) name
#define all(box) box.begin(), box.end()

template <class T, class U> inline bool chmax(T& lhs, const U& rhs) { if(lhs < rhs) { lhs = rhs; return 1; } return 0; }
template <class T, class U> inline bool chmin(T& lhs, const U& rhs) { if(lhs > rhs) { lhs = rhs; return 1; } return 0; }

int N;
int A[200000];

int main() {

    cin >> N;
    rep(i, N) cin >> A[i];

    int now = 0;
    rep(i, N) {
        if(A[i] == now + 1) ++now;
    }

    if(now == 0) cout << -1 << '\n';
    else cout << N - now << '\n';

}