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;

fn read_line() -> String {
    let mut s = String::new();
    io::stdin().read_line(&mut s).unwrap();
    s
}

macro_rules! from_line {
    ($($a:ident : $t:ty),+) => {
        $(let $a: $t;)+
        {
            let _line = read_line();
            let mut _it = _line.trim().split_whitespace();
            $($a = _it.next().unwrap().parse().unwrap();)+
                assert!(_it.next().is_none());
        }
    };
}

fn main() {
    from_line!(n: i64);

    let mut d: Vec<i64> = vec![0];
    let mut d_clone: Vec<i64>;
    for i in 0..n {
        d_clone = d.clone();
        let a: i64;
        from_line!(a: i64);
        let mut have = false;

        for j in d_clone {
            if j == a {
                have = true;
                break;
            }
        }
        if have == false {
            d.push(a);
        }
    }

    println!("{}", d.len() - 1);
}