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>
#define rep(i, n) for (ll i = 0; i < n; i++)
#define FOR(i, m, n) for(ll i = m;i < n;i++)
#define VSORT(v) sort(v.begin(), v.end());
#define VREV(v) reverse(v.begin(), v.end());
using namespace std;
using ll = long long;
using pll = pair<ll,ll>;
using vll = vector<ll>;
using vdouble = vector<double>;
using vstr = vector<string>;
typedef pair<int,int> P;
int dx[4] = {1, 0, -1, 0},dy[4] = {0, 1, 0, -1};
ll gcd(ll a, ll b) { return b?gcd(b,a%b):a;}
ll lcm(ll a, ll b) { return a/gcd(a,b)*b;}
template<class T> inline bool chmax(T& a, T b){if(a<b) {a=b; return 1;} return 0; }
template<class T> inline bool chmin(T& a, T b){if(a>b) {a=b; return 1;} return 0; }
const ll INF = 1LL << 60;
const ll mod = 1000000007;
int main(){
	//乱数srand((unsigned)time(NULL));
	ll N; cin>>N;
	vll a(N); rep(i,N) cin>>a[i];
	ll mans=1;
	rep(i,N){
		if(mans==a[i]) mans++;
	}
    if(mans==1) cout<<-1<<endl;
	else cout<<N-mans+1<<endl;
}