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: Runtime Error)

import java.util.Arrays;
import java.util.Comparator;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		Integer[] mochi = new Integer[N];
		int ans = 1;
		
		for(int i =0; i< N; i++) {
			mochi[i] = sc.nextInt();
		}
		Arrays.sort(mochi,Comparator.reverseOrder());
		
		for(int i = 0; i < N; i++) {
			if(i != N - 1) {
				if(mochi[i] > mochi[i+1]) {
					ans++;
					}
				} else {
				if(mochi[i] != mochi[i-1]) {
					ans++;
				}
			}
		}
		System.out.println(ans);
	}
}