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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;

#define ll long long
#define fast ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define endl "\n"
#define pb push_back
#define all(a) (a).begin(),(a).end()
#define rev(a) (a).rbegin(),(a).rend()
#define pll pair<ll,ll>
#define rep(i,a,b) for(long long i = (a); i <= (b); i++)

const int mod = 1e9+7;
const int N = 1e5+5; 

int main(){
	fast;
	ll n; cin >> n;
	ll arr[n+5];
	rep(i,1,n) cin >> arr[i];
	
	ll ans = 0;
	ll curr = 1;
	
	rep(i,1,n){
	    if(arr[i] == curr){
	        curr++;
	    }
	    else{
	        ans++;
	    }
	}
	
	if(ans == n)
	cout << ans << endl;
	else
	cout << "-1\n";
}