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: Time Limit Exceeded)

import java.util.*;
import java.util.stream.IntStream;
import java.util.concurrent.TimeUnit;
import java.util.concurrent.CountDownLatch;

public class Main{
    
    static final int SLEEP_TIME = 18;
    
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        
        int n = Integer.parseInt(sc.next());
        int[] d = new int[n];
        for(int i=0; i<n; i++){
            d[i] = Integer.parseInt(sc.next());
        }
        
        final CountDownLatch latch = new CountDownLatch(d.length+1);
        final Count c = new Count();
        
        new Thread(new Runnable(){
            public void run(){
                try{
                    for(int i=0; i<120; i++){
                        TimeUnit.MILLISECONDS.sleep(SLEEP_TIME);
                        synchronized(c){
                            c.countUp();
                        }
                    }
                }catch(InterruptedException e){
                    e.printStackTrace();
                }finally{
                    latch.countDown();
                }
            }
        }).start();
        
        for(final int num : d){
            new Thread(new Runnable(){
                public void run(){
                    try{
                        TimeUnit.MILLISECONDS.sleep(num*SLEEP_TIME);
                        synchronized(c){
                            c.flagUp();
                        }
                    }catch(InterruptedException e){
                        e.printStackTrace();
                    }finally{
                        latch.countDown();
                    }
                }
            }).start();
        }
        
        try{
            latch.await();
        }catch(InterruptedException e){
            e.printStackTrace();
        }
        
        System.out.println(c.getResult());
    }
}

class Count{
    int cnt;
    boolean flag;
    
    public Count(){
        cnt = 0;
        flag = false;
    }
    
    public void flagUp(){
        flag = true;
    }
    
    public void countUp(){
        if(flag){
            cnt++;
            flag = false;
        }
    }
    
    public int getResult(){
        return cnt;
    }
}