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++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
using ll = long long;

#define debug(x) (cerr << #x << " = " << x << endl;)
#define all(v) (v).begin(), (v).end()
#define rall(v) (v).rbegin(), (v).rend()
#define FOR(it, b, e) for (typeof(b) it = (b); it != (e); ++it)
#define MSET(c, v) memset(c, v, sizeof(c))

const int INF = 0x3F3F3F3F; 
const int NEGINF = 0xC0C0C0C0;
const int NULO = -1;
const ll mod = 1e9 + 7;
const double EPS = 1e-10;

inline int cmp(double x, double y = 0) 
{  
  if(fabs(x - y) < EPS)
    return 0;
  return x > y ? 1 : -1;
}

const int maxn = 2e5 + 100;

ll a[maxn];

ll exp(ll a, ll b)
{
  ll ans = 1;
  while(b)
  {
    if(b & 1)
      ans = (ans * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return ans;
}

ll lcm(ll a, ll b)
{
  return a * (b / __gcd(a,b));
}

char f(char c)
{
  if(c == 'r')
    return 'p';
  else if(c == 'p')
    return 's';
  else
    return 'r';
}

int main()
{

  ios::sync_with_stdio(0);
  int n;
  cin >> n;
  int j = 1;
  for(int i = 0 ; i < n ; i++)
  {
    int a;
    cin >> a;
    if(a == j)
      j++;
  }

  cout << ( j == 1 ? -1 : (n - j + 1)) << endl;
  return 0;
}