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 <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <functional>
#include <queue>
#include <string>
#include <cstring>
#include <numeric>
#include <cstdlib>
#include <cmath>
#include <map>
#include <unordered_map>
#include <set>
using namespace std;

typedef long long ll;

#define INF 10e17
#define rep(i,n) for(long long i=0; i<n; i++)
#define repr(i,n,m) for(long long i=m; i<n; i++)
#define END cout << endl
#define MOD 1000000007
#define pb push_back
#define sorti(x) sort(x.begin(), x.end())
#define sortd(x) sort(x.begin(), x.end(), std::greater<long long>())
#define debug(x) std::cerr << (x) << std::endl;
#define roll(x) for (auto&& itr : x) { debug(itr); }

template <class T> inline void chmax(T &ans, T t) { if (t > ans) ans = t;}
template <class T> inline void chmin(T &ans, T t) { if (t < ans) ans = t;}

int main() {
  int n;
  cin >> n;
  int t = -1, now = 0;
  map<int, set<int>> mp;
  rep(i, n) {
    int l;
    cin >> l;
    if (l == now + 1) {
      now = l;
    }
  }
  if (now == 0) cout << -1 << endl;
  else
  cout << n - now << endl;
}