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;

fn main() {
    let n = parse_line::<usize>();
    let line = parse_line::<String>();
    let split: Vec<&str> = line.split_whitespace().collect();
    let nums: Vec<usize> = split.iter().map(|s| s.parse::<usize>().unwrap()).collect();
    let found = find_numbers(nums);
    if found == 0 {
        println!("{}", -1);
    } else {
        println!("{}", n - found);
    };
}

fn parse_line<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn find_numbers(nv: Vec<usize>) -> usize {
    let mut counter = 1;
    for i in nv {
        if i == counter {
            counter += 1;
        }
    }
    counter - 1
}