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<iostream>
#include<string>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cmath>
#include<vector>
#include<utility>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<unordered_set>
#include<unordered_map>
#include<bitset>
using namespace std;
#define INF 2000000007
#define LINF 100000000000000007
#define MOD 1000000007
#define int long long
#define rep(i,n) for(int i=0;i<n;i++)
#define repb(i, n) for (int i = n - 1; i >= 0; i--)
#define MODE 0
#ifdef MODE
#define DEB(X) cout<< #X <<": "<<X<<" ";
#define ARDEB(i,X) cout<< #X <<"["<<i<<"]: "<<X[i]<<" ";
#define END cout<<endl;
#else
#define DEB(X) {}
#define ARDEB(i,X) {}
#define END {}
#endif
typedef long long ll;
typedef pair<int,int> P;
struct edge{int to,cost;};
int ceil2(int a,int b){if(a%b){return a/b+1;}else{return a/b;}}
using namespace std;

int n,ans;
int a[222222];
signed main(){
	cin>>n;
	rep(i,n)cin>>a[i];
	int now=1;
	rep(i,n){
		if(a[i]==now){
			now++;
		}
	}
	cout<<n-now+1<<endl;
}