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: Wrong Answer)

import java.util.Scanner;
public class Main{
    public static void main(String[]args){
        Scanner sc=new Scanner(System.in);
        int N=sc.nextInt();
        int a;
        int x=0;
        int [] b=new int[N];
        //i番目の数の入力を行う
        for(int i=0;i<=N-1;i++){
             a=sc.nextInt();
             b[i]=a;  
        }
        int l=0;
        for(int j=0;j<=N-1;j++){
            for(int k=j+1;k<=N-1;k++){
                if(b[j]==b[k]){
                    l=l+1; 
                    for(int f=k+1;f<=N-1;f++){
                        if(b[j]==b[f]){
                            x=x+1;
                        }
                        }
                        
                }
                

            }
        if(j==N-1){
            System.out.println(N-l+x);
        }
        }

        


    }
}