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::collections::HashSet;

fn procon_read<T: std::str::FromStr>() ->  Vec<T> {
	let mut buf = String::new();
	io::stdin().read_line(&mut buf).expect("");
	let iter = buf.split_whitespace();
	let mut v: Vec<T> = Vec::new();
	for i in iter {
		v.push(i.parse().ok().unwrap());
	}
	v
}

fn main() {
	let v:Vec<i32> = procon_read();
	let n = v[0];
	let mut d:Vec<i32> = Vec::new();
	for i in 0..n {
		let v:Vec<i32> = procon_read();
		d.push(v[0]);
	}
	d.sort();
	let mut set = HashSet::new();
	for i in d {
		set.insert(i);
	}
	println!("{}", set.len());
}