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)

/*by freesteed*/
#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define pb push_back
#define fill(x,c) memset(x,c,sizeof(x))
#define fi first
#define se second
#define sz(A) (ll)(A.size())
#define all(x) (x).begin(),(x).end()
#define caset int ___T; cin>>___T; for(int cs=1;cs<=___T;cs++)
#define get_be(c,x) (lower_bound(c.begin(),c.end(),x)-c.begin())
//get_be biger & equal
typedef long long ll;
typedef pair<int,int> PII;
typedef vector<int> VI;
const ll mod=1e9+7;
const ll oo=1e9+10;
const int Max = 1e6+10;
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll n,k,r,S,p,ans;
string s; 
map<char,int> score;
map<char,int> id;
int help(vector<int>& slices) {
	 int ans=0;
	 int preWin=0;
	 rep(i,0,sz(slices)){
	 	if(!preWin){
	 		preWin = 1;
	 		ans+=score[s[slices[i]]];
		 }else{
		 	if(s[slices[i]]!=s[slices[i-1]]){
		 		ans+=score[s[slices[i]]];
			 }else{
			 	preWin = 0;
			 }
		 }
	 }
	 return ans;
} 
void solve(){
	ll n;
	cin>>n;
	ll need = 1;
	rep(i,0,n){
		ll t;
		cin>>t;
		if(t==need)need++;
	}
	if(need==1){
		cout<<-1<<endl;
	}else{
		cout<<n-(need-1);
	}
}
int main() {
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	//caset{
		solve();
	//}
  	return 0;
}