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)

//****** @author mdazmat9 **********
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define UB upper_bound
#define LB lower_bound
#define BS binary_search
#define MP make_pair
#define EB emplace_back
#define endl "\n"
#define PI 3.1415926535897932384626433832795
#define MOD 1000000007
#define F first
#define S second
#define ALL(a) (a).begin(),(a).end()
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
ll fast_pow(ll x, ll y, ll p)
{
    ll res = 1;
    x = x % p;
    while (y > 0)
    {
        if (y & 1)
            res = (res*x) % p;
        y = y>>1;
        x = (x*x) % p;
    }
    return res;
}



void solve() {
   ll n;cin>>n;
   ll a[n];
   for(ll i=0;i<n;i++)cin>>a[i];
   ll x=-1;
   int c=1;
   for(ll i=0;i<n;i++){
       if(a[i]==c){
           x=c;
           c++;
       }
   }
   if(x==-1)cout<<-1<<endl;
   else cout<<n-x<<endl;

}
int main()
{
    IOS;
    ll t=1;
   //   cin>>t;
    while(t--){
        solve();
    }
    return 0;
}