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 Rust (1.15.1) (judge verdict: Accepted)

use std::io::{self, Read};

fn main() {
	let mut input = String::new();
	let _ = io::stdin().read_to_string(&mut input);
	let lines: Vec<&str> = input.split("\n").collect();	
	let N = lines[0].parse::<usize>().unwrap_or(0);
	let a: Vec<i64> = lines[1].split(" ").map(|x| x.parse::<i64>().unwrap_or(0)).collect();
	let mut ans = 0;
	let mut cursor = 1;
	for i in 0..N {
		if a[i] != cursor {
			ans += 1;
		} else {
			cursor += 1;
		}
	}
	if ans == N {
		println!("{}", -1);
	} else {
		println!("{}", ans);
	}
}