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>

#define REP(i, x, n) for (int i = x; i < (int)(n); i++)
#define rep(i, n) REP(i, 0, n)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define sz(x) (int)(x.size())
#define popcount(x) __builtin_popcount(x)
#define popcountll(x) __builtin_popcountll(x)
#define uniq(x) x.erase(unique(x.begin(), x.end()), x.end())
#define F first
#define S second
#define mp make_pair
#define eb emplace_back
#define pii pair<int, int>
#define pll pair<ll, ll>

using namespace std;

typedef long long ll;

/* --- INFやMODの値が適切か確認する --- */
// const int INF = 1 << 30;
// const ll INF = 1LL << 60;
// const ll MOD = 1000000007;

int main() {
  int n;
  cin >> n;
  vector<int> a(n);
  rep(i, n) cin >> a[i];
  int t = 0;
  rep(i, n) if (t + 1 == a[i]) { t++; }
  if (t == 0)
    cout << "-1" << endl;
  else
    cout << n - t << endl;
}