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>


#define rep2(x,fr,to) for(int (x)=(fr);(x)<(to);(x)++)
#define rep(x,to) for(int (x)=0;(x)<(to);(x)++)
#define repr(x,fr,to) for(int (x)=(fr);(x)>=(to);(x)--)
#define all(c) (c).begin(),(c).end()
#define sz(v) (int)(v).size()

using namespace std;
typedef long long ll; typedef vector<int> VI; typedef pair<int,int> pii;
const int MD = (int)1e9 + 7; typedef vector<ll> VL;
void dbg(){ cerr << "\n"; }  template <class T, class ...T2>
void dbg(const T& fst, const T2&...rst){ cerr << fst << ": "; dbg(rst...); }
template <class T> void amax(T& a,T b){ a=max(a,b);}

	ll gcd(ll x, ll y){ return y? gcd(y, x%y): x; }
	ll lcm(ll x, ll y){ return (x  / gcd(x, y) * y ); }

int main() {
	//cin.tie(0); ios_base::sync_with_stdio(false);
	int n;
	cin >>n;
	VI a(n), gm, ck(n+1);
	rep(i, n){ cin >>a[i]; a[i]--;}
	ll ans =0;
	int p= 0;
	rep(i, n){
		if(p != a[i]){
			ans++;
		}else p++;
		
	}
	
	cout << (ans != n? ans: -1) <<"\n";
	return 0;
}