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)

// Etavioxy
#include<cstdio>
#include<cctype>
#include<cstring>
#include<algorithm>
#include<cmath>
#define il inline
#define ll long long
#define rep(i,s,t) for(register int i=(s);i<=(t);i++)
#define rev_rep(i,s,t) for(register int i=(s);i>=(t);i--)
#define each(i,u) for(int i=head[u];i;i=bow[i].nxt)
#define file(s) freopen(s".in" ,"r",stdin),freopen(s".out","w",stdout)
#define pt(x) putchar(x)
using namespace std;
il int ci(){
	register char ch;int f=1;
	while(!isdigit(ch=getchar()))f=ch=='-'?-1:1;
	register int x=ch^'0';
	while(isdigit(ch=getchar()))x=(x*10)+(ch^'0');
	return f*x;
}

int main(){
	int n=ci();
	int now=1, ans=0;
	rep(i,1,n){
		int x=ci();
		if( x==now ) now++;
		else ans++;
	}
	if( now==1 ) puts("-1");
	else printf("%d\n",ans);
	return 0;
}