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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.*;
class Main{
  public static void main(String[] args){
    Scanner sc = new Scanner(System.in);
    int n = Integer.parseInt(sc.next());
    Integer[] d = new Integer[n];
    for (int i = 0; i < n; i++){
      d[i] = Integer.parseInt(sc.next());
      // 右辺はint型だがオートボクシングされている
    }
    Arrays.sort(d,(n1,n2) -> {return n2 - n1;});
    int ans = 0, last = 101;
    for (int i = 0; i < n; i ++){
      if (last != d[i]){
        ans += 1;
        last = d[i];
      }
    }
    System.out.println(ans);
    
  }
}