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;
using ll = long long;
using ld = long double;
using str = string;
typedef vector<string> vs;
typedef vector<bool> vb;
typedef vector<char> vc;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<vs> vvs;
typedef vector<vc> vvc;
typedef vector<vb> vvb;
using P = pair<ll,ll>;
#define REP(i, n) for (ll i = 0; i < (int)(n); i++)
#define RREP(i, n) for (ll i = (int)(n)-1; i > -1 ; i--)
#define IN(T, x) T x;cin >> x;
#define AIN(T, a, n) vector<T> a(n);REP(i, n){cin >> a[i];}
#define A2IN(T1, a, T2, b, n) vector<T1> a(n);vector<T2> b(n);REP(i, n){cin >> a[i] >> b[i];}
#define ALL(a) (a).begin(),(a).end()
#define SORT(a) sort(ALL(a))
#define RSORT(a) SORT(a);reverse(ALL(a))
#define pb push_back
#define mp make_pair
#define PF first
#define PS second
template<class T>bool chmax(T &a,const T &b){if(a<b){a=b;return 1;}return 0;}
template<class T>bool chmin(T &a,const T &b){if(b<a){a=b;return 1;}return 0;}
template<typename A, size_t N, typename T>void Fill(A (&array)[N], const T &val){fill( (T*)array, (T*)(array+N), val );}
#define OUT(x) cout << x << endl;
#define OUTD(x) cout << fixed << x << endl;
#define BOUT(b) if(b){OUT("Yes");}else{OUT("No");}
const ll MOD = 1000000007;
const ll INF = 10e9;
#define PI 3.14159265369;
ll dx[4] = {1,-1,0,0};
ll dy[4] = {0,0,1,-1};
ll ddx[8] = {1,1,1,-1,-1,-1,0,0};
ll ddy[8] = {0,1,-1,0,1,-1,1,-1};

ll dp[1001][1001];

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  
  IN(ll,N);
  AIN(ll,A,N);
  ll count = 1;
  REP(i,N){
    if(A.at(i) == count)
      count++;
  }
  if(count == 1){
    OUT(-1);
  }else{
    OUT(N - count + 1);
  }
  
}