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)

/*
 ABC148 D
 auther: takeheroaf
 created at 2020-1-10 10:10:21
*/

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, j, n) for (int i = j; i < n; i++)
#define all(x) (x).begin(),(x).end()
#define pii pair<int, int>
#define pll pair<ll, ll>
#define vc vector
#define a first
#define b second
#define pb push_back
#define INF (1000000000)
#define MOD (1000000007)
#define MAX (100000)

/////////////////////////////////////////////////////////

template<typename T1, typename T2>
void chmin(T1 &a, T2 b) {if(a > b) a = b;}
template<typename T1, typename T2>
void chmax(T1 &a, T2 b) {if(a < b) a = b;}

template<typename T1, typename T2>
ll mypow(T1 a, T2 n){
  if(n == 0) return 1;
  if(n == 1) return a;
  if(n % 2) return a * mypow(a, n - 1);
  ll tmp = mypow(a, n / 2);
  return tmp * tmp;
}

ll comb(ll n, ll r){
	ll res = 1;
  rep(i, 0, r){
    res *= n - i;
    res /= i + 1;
  }
  return res;
}

/////////////////////////////////////////////////////////

void Main(){
  int N; cin >> N;
  vc<int> A(N);
  rep(i, 0, N) cin >> A[i];
  int cur = 1;
  rep(i, 0, N){
    if(A[i] == cur){
      cur++;
    }
  }
  if(cur == 1){
    cout << -1 << endl;
  }else{
    cout << N - (cur - 1) << endl;
  }
  return;
}

/////////////////////////////////////////////////////////

int main(){
  cin.tie(nullptr);
  ios_base::sync_with_stdio(false);
  cout << std::fixed << std::setprecision(15);

  Main();

  return 0;
}