Problem:
Score : 200 points

Problem Statement

An X-layered kagami mochi (X ≥ 1) is a pile of X round mochi (rice cake) stacked vertically where each mochi (except the bottom one) has a smaller diameter than that of the mochi directly below it. For example, if you stack three mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, you have a 3-layered kagami mochi; if you put just one mochi, you have a 1-layered kagami mochi.

Lunlun the dachshund has N round mochi, and the diameter of the i-th mochi is d_i centimeters. When we make a kagami mochi using some or all of them, at most how many layers can our kagami mochi have?

Constraints

1 ≤ N ≤ 100

1 ≤ d_i ≤ 100

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
d_1
:
d_N

Output

Print the maximum number of layers in a kagami mochi that can be made.

Sample Input 1

4
10
8
8
6

Sample Output 1

3

If we stack the mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, we have a 3-layered kagami mochi, which is the maximum number of layers.

Sample Input 2

3
15
15
15

Sample Output 2

1

When all the mochi have the same diameter, we can only have a 1-layered kagami mochi.

Sample Input 3

7
50
30
50
100
50
80
30

Sample Output 3

4

Solution in Rust (1.15.1) (judge verdict: Accepted)

use std::io::*;
use std::str::FromStr;

struct Scanner<R: Read> {
    reader: R,
}

#[allow(dead_code)]
impl<R: Read> Scanner<R> {
    fn new(reader: R) -> Scanner<R> {
        Scanner { reader: reader }
    }

    fn safe_read<T: FromStr>(&mut self) -> Option<T> {
        let token = self.reader.by_ref().bytes().map(|c| c.unwrap() as char)
            .skip_while(|c| c.is_whitespace())
            .take_while(|c| !c.is_whitespace())
            .collect::<String>();
        if token.is_empty() {
            None
        } else {
            token.parse::<T>().ok()
        }
    }

    fn read<T: FromStr>(&mut self) -> T {
        if let Some(s) = self.safe_read() {
            s
        } else {
            writeln!(stderr(), "Process terminated with EOF").unwrap();
            std::process::exit(0);
        }
    }
}

fn main() {
    std::thread::Builder::new()
        .stack_size(104_857_600)
        .spawn(solve)
        .unwrap()
        .join()
        .unwrap();
}

fn solve() {
    let cin = stdin();
    let cin = cin.lock();
    let mut sc = Scanner::new(cin);
    loop {
        let n: usize = sc.read();
        let mut d: Vec<i32> = (0..n).map(|_| sc.read()).collect();
        d.sort();
        d.dedup();
        println!("{}", d.len());
    }
}