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(unused_imports)]
#![allow(dead_code)]

use std::io::{ self, BufRead };
use std::iter::Iterator;
use std::cmp::*;
use std::collections::*;

fn sum(mut x: i64) -> i64 {
    let mut y = 0;
    while x > 0 {
        y += x % 10;
        x /= 10;
    }
    y
}

fn main() {
    let stdin = io::stdin();
    let mut cin = stdin.lock();

    let mut it = cin.lines()
    .map(|s| s.unwrap().parse::<i64>().unwrap());
    let n = it.next().unwrap();

    let mut d = it.take(n as usize).collect::<Vec<_>>();
    d.sort();
    let mut ans = 1;
    for i in 1..d.len() {
        if d[i-1] != d[i] { ans += 1; }
    }    

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