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<iostream>
#include<vector>
#include<string>
#include<algorithm>
#include<cmath>
#include<bitset>
#include<deque>
#include<functional>
#include<iterator>
#include<map>
#include<set>
#include<stack>
#include<queue>
#include<utility>
using namespace std;
typedef long long ll;
#define a first
#define b second
#define sz(x) (ll)((x).size())
#define pb push_back
#define mp make_pair
#define bg begin()
#define ed end()
#define all(x) (x).bg,(x).ed
#define rep(i,n) for(ll i=0;i<(n);i++)
#define rep1(i,n) for(ll i=1;i<=(n);i++)
#define rrep(i,n) for(ll i=(n)-1;i>=0;i--)
#define rrep1(i,n) for(ll i=(n);i>=1;i--)
#define FOR(i,a,b) for(ll i=(a);i<(b);i++)
const ll MOD=1000000007;
const ll INF=1000000000000000;
template<class T> inline bool chmin(T& a, T b){if(a>b){a=b;return true;}return false;}
template<class T> inline bool chmax(T& a, T b){if(a<b){a=b;return true;}return false;}
ll maxx(ll x,ll y,ll z){return max(max(x,y),z);}
ll minn(ll x,ll y,ll z){return min(min(x,y),z);}
ll gcd(ll x,ll y){if(x%y==0) return y;else return gcd(y,x%y);}
ll lcm(ll x,ll y){return x*(y/gcd(x,y));}
ll digsz(ll x){if(x==0) return 1;else{ll ans=0;while(x){x/=10;ans++;}return ans;}}
ll digsum(ll x){ll sum=0;while(x){sum+=x%10;x/=10;}return sum;}
vector<ll> pw2(62,1);vector<ll> pw10(19,1);

int main(){
    {rep1(i,61) pw2[i]=2*pw2[i-1];}
    {rep1(i,18) pw10[i]=10*pw10[i-1];}
    
    ll N; cin>>N;
    vector<ll> A(N);
    rep(i,N) cin>>A[i];
    
    ll now=1;
    rep(i,N){
        if(A[i]==now) now++;
    }
    
    if(now==1) cout<<-1;
    else cout<<N-(now-1);
}