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;

pub fn read<T: FromStr>() -> T {
  let stdin = stdin();
  let stdin = stdin.lock();
  let token: String = stdin
    .bytes()
    .map(|c| c.expect("failed to read char") as char)
    .skip_while(|c| c.is_whitespace())
    .take_while(|c| !c.is_whitespace())
    .collect();
  token.parse().ok().expect("failed to parse token")
}

use std::cmp::{max, min};
use std::collections::HashSet;

fn main() {
  let n = read::<i64>();
  let mut d = HashSet::new();

  for _ in 0..n {
    d.insert(read::<i64>());
  }

  println!("{}", d.len());

}