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)

#pragma GCC optimize ("O3")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <numeric>
#include <vector>
#include <set>
#include <map>
#include <cstring>
#include <stack>
#include <unordered_map>
#define lld long long int
#define dbg1(vari1) cerr<<#vari1<<" = "<<(vari1)<<endl;
#define dbg2(vari1,vari2) cerr<<#vari1<<" = "<<(vari1)<<" "<<#vari2<<" = "<<(vari2)<<endl;
#define dbg3(vari1,vari2,vari3) cerr<<#vari1<<" = "<<(vari1)<<" "<<#vari2<<" = "<<(vari2)<<" "<<#vari3<<" = "<<(vari3)<<endl;
using namespace std;
int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	int n;
	cin >> n;
	int arr[n];
	for(int i=0;i<n;i++)
		cin >> arr[i];
	int ch=1;
	for(int i=0;i<n;i++)
	{
		if(arr[i]==ch)
			ch++;
	}
	if(ch==n+1)
		cout << "0";
	else if(ch==1)
		cout << "-1";
	else
		cout << n-ch+1;
}