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 (GCC 5.4.1) (judge verdict: Accepted)

char buf[2000000];
unsigned sz;
char out[9];
main()
{
	read(0,buf,2000000);
	unsigned N=0;
	for(;;)
	{
		char k=buf[sz++];
		if(k<'0')break;
		N=N*10+k-'0';
	}
	unsigned now=1;
	for(unsigned i=0;i<N;i++)
	{
		unsigned A=0;
		for(;;)
		{
			char k=buf[sz++];
			if(k<'0')break;
			A=A*10+k-'0';
		}
		now+=A==now;
	}
	out[8]='\n';
	unsigned outsz=8;
	if(now==1)
	{
		out[7]='1';
		out[6]='-';
		outsz=6;
	}
	else
	{
		N-=now-1;
		do{
			out[--outsz]=N%10+'0';
		}while(N/=10);
	}
	write(1,out+outsz,9-outsz);
}