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;

  const int MOD = 1000000007;

  //long long
  using ll = long long;

  //出力系
  #define print(x) cout << x << endl
  #define yes cout << "Yes" << endl
  #define YES cout << "YES" << endl
  #define no cout << "No" << endl
  #define NO cout << "NO" << endl
  
  // begin() end()
  #define all(x) (x).begin(),(x).end()

  //for
  #define REP(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)

  //最大公約数（ll）
  ll gcd(ll x, ll y) { return y ? gcd(y,x%y) : x;}

  //素因数分解
  vector<pair<ll, int>>factorize(ll n){
    vector<pair<ll, int>> res;
    for(ll i = 2; i * i <= n; ++i){
      if(n % i) continue;
      res.emplace_back(i, 0);
      while(n % i == 0){
        n /= i;
        res.back().second++;
      }
    }
  if(n != 1) res.emplace_back(n, 1);
  return res;
  }


int main()
{
    int N;
    cin >> N;
    vector<int>A(N);
    REP(i, N) cin >> A.at(i);
    int target = 1;
    int cnt = 0;
    REP(i, N){
        if(A.at(i) == target){
            target++;
            cnt++;
        }
    }
    if(cnt != 0) print(N - cnt);
    else print(-1);
    }