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)

#![allow(non_snake_case,unused_imports)]
use std::io::Read;
use std::cmp::max;
use std::cmp::min;
use std::str::FromStr;
use std::collections::HashMap;
use std::collections::HashSet;
use std::collections::BinaryHeap;

fn main() {
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();
    let mut iter = buf.split_whitespace();
    let N:usize = iter.next().unwrap().parse().unwrap();
    let mut D = HashSet::new();
    let _ = (0..N).map(|_| D.insert(iter.next().unwrap().parse::<usize>().unwrap())).collect::<Vec<_>>();

    let ans = D.len();

    println!("{}",ans);
}