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;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<long long,long long> pll;
#define ep emplace_back
#define pb push_back
#define mp make_pair
#define rep(i,n) for(int i=0;i<(n);++i)
#define all(v) v.begin(), v.end()
#define revall(v) v.rbegin(), v.rend()
constexpr int mod=1000000007;
constexpr int mod1=998244353;
vector<int> dx={0,1,0,-1},dy={-1,0,1,0};
bool inside(int y,int x,int h,int w){ if(y<h && y>=0 && x<w && x>=0) return true; return false; }
template<class T> inline bool chmin(T& a, T b){ if(a > b){ a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b){ if(a < b){ a = b; return true; } return false; }






int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	int n;cin >> n;
	vector<int> a(n);
	rep(i,n) cin >> a.at(i);
	int k = 1;
	rep(i,n){
		if(a.at(i) == k) k++;
	}
	if(k == 1) cout << -1 << endl;
	else cout << n - k + 1 << endl;
}