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::*;

fn read <T: FromStr>(s: &mut StdinLock) -> Option<T> {
    let s = s.by_ref().bytes().map(|c| c.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect::<String>();
    s.parse::<T>().ok()
}

pub fn main() {
    let stdin = stdin();
    let mut stdin = stdin.lock();

    let n: u32 = read(&mut stdin).unwrap();
    let nums: Vec<u32> = (0..n).map(|_x| read(&mut stdin).unwrap()).collect();

    // let mut res: Vec<u32> = Vec::with_capacity(n as usize);
    // for i in 0..n as usize {
    //     let x = nums[i];

    //     if !res.contains(&x) {
    //         res.push(x);
    //     }
    // }

    // println!("{:?}", res.len());

    let mut res: [u32; 100] = [0; 100];
    for num in nums {
        let num = (num - 1) as usize;
        res[num] += 1;
    }
    let result = res.iter().filter(|x| x > &&0).fold(0, |acc, _x| acc + 1);
    println!("{:?}", result);
}