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;
typedef		 long long int    ll;
typedef		vector<ll>      vll;
#define     fi              first
#define     print(v)        for(ll i:v) cout<<i<<ss
#define     se              second
#define		pb              push_back
#define		nn              "\n"
#define		all(p)          p.begin(),p.end()
#define		zz(v)           (ll)v.size()
#define 	S(a)            scanf("%lld",&a)
#define 	SS(a,b)         scanf("%lld %lld",&a,&b)
#define 	SSS(a,b,c)      scanf("%lld %lld %lld",&a,&b,&c)
#define		ss              ' '
#define     pii             pair<ll,ll>
#define		gcd(a,b)        __gcd(a,b)
#define		lcm(a,b)        (a*b)/gcd(a,b)
const double eps = 1e-8;



int main()
{

    ll n;
    cin>>n;
    ll ar[n+10];
    for(ll i=0; i<n; i++)
    {
        cin>>ar[i];
    }
    ll ans=n;
    ll c=0;
    ll i=0;
    ll k=1;
    while(i<n)
    {
        if(ar[i]==k)
        {
            c++;
            k++;
        }
        i++;
    }
    if(!c)
        cout<<-1<<nn;
    else
        cout<<n-c<<nn;



}


/* Final check before submit :

1. array size or integer overflow,like uninitialised 0 index.

2. Think twice,code once.check all possible counter test case.

3. Be careful about corner case! n=1?k=1? something about 0?

4. avoid stupid mistake!complexity(t/m)?precision error?submit same code twice?

5. if got WA than remember that you are missing something common.
   *** Be confident!!! who knows? may be your one step back to AC ***
4. minus mod ans=(ans-k+mod)%mod;

*/