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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.HashSet;
import java.util.Scanner;
import java.util.Set;
import java.util.stream.Stream;

public class Main
{
    private static int f(int[] xs, int pos, int max, int count) {
        if (pos >= xs.length) {
            return count;
        }

        if (xs[pos] < max) {
            return Math.max(
                    f(xs, pos + 1, xs[pos], count + 1),
                    f(xs, pos + 1, max, count));
        }
        else {
            return f(xs, pos + 1, max, count);
        }
    }

    public static void main(String[] args)
    {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] xs = new int[n];
        for (int i = 0; i < n; i++) {
            xs[i] = sc.nextInt();
        }

//        System.out.println(f(xs, 0, Integer.MAX_VALUE, 0));
        Set<Integer> set = new HashSet<>();
        for (int i : xs) {
            set.add(i);
        }
        System.out.println(set.size());
   }
}