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;
#define int long long
#define itn int
#define mod 1000000007
#define INFi 2147483647
#define INFl 9223372036854775807LL
#define fi first
#define se second
#define be begin()
#define en end()
int abs(int a,int b){
    return max(a,b)-min(a,b);
}
int gcd(int a,int b){//最大公約数
int c=a%b;
while(true){
if(c==0) break;
a=b;
b=c;
c=a%b;
}
return b;
}
bool IsPrime(int num){
if (num < 2) return false;
else if (num == 2) return true;
else if (num % 2 == 0) return false;
double sqrtNum = sqrt(num);
for(int i=3;i<=sqrtNum;i+=2){
if(num%i==0){
return false;
}
}
}
int lcm(int a,int b){//最小公倍数
return (a*b)/gcd(a,b);
}	
int q[10000000];
int w[10000000];
signed main(){
// __builtin_ctz(N); Nを２で何回割り切れるかを返す
//(n+1)*n/2; １からNまでの総和を求める式
    cin.tie(0);
    ios::sync_with_stdio(false);
    int a,b=0,c=0;
    cin>>a;
    for(int i=0;i<a;i++) cin>>q[i];
    for(int i=0;i<a;i++){
	if(q[i]!=i+1-b) b++;
    }
    if(b!=a) cout<<b<<"\n";
    else cout<<"-1\n";
}