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>
using namespace std;
using ll = long long;
using vl = vector<ll>;
using vvl = vector<vl>;
using vi = vector<int>;
using vvi = vector<vi>;
#define ote(x) cout<<(x)<<endl
#define all(x) (x).begin(),(x).end()
#define rp(i,s,e) for(int i=(s);i<(e);++i)
#define rpz(i,e) rp(i,0,e)

ll solve(int N, vi &A){
	ll ans = 0;
	int ix = 1;
	rp(i,0,N){
		if(A[i]!=ix){
			ans++;
		}
		else{
			ix++;
		}
	}
	return ans<N?ans:-1;
}

int main(){
	int N; cin>>N;
	vi A(N);
	rp(i,0,N)cin>>A[i];
	ote(solve(N,A));
}