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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;

public class Main {
	final int _intMax = Integer.MAX_VALUE; // =2147483647>10^9
	final int _intMin = Integer.MIN_VALUE;
	final long _longMax = Long.MAX_VALUE; // =9223372036854775807L>10^18
	final long _longMin = Long.MIN_VALUE;
	static boolean bElapsed = false;

	void solve() {
		int n = readNum();
		int[] ia = new int[n];
		for (int i = 0; i < n; i++) {
			ia[i] = readNum();
		}
		Arrays.sort(ia);
		int ans = 1;
		for (int i = 1; i < n; i++) {
			if (ia[i - 1] != ia[i]) {
				ans++;
			}
		}
		pln(ans);
	}

	int pint(String s) {
		return Integer.parseInt(s);
	}

	long plong(String s) {
		return Long.parseLong(s);
	}

	String readLine() {
		try {
			return _in.readLine();
		} catch (IOException e) {
			throw new RuntimeException(e);
		}
	}

	int readNum() {
		String line = readLine();
		return pint(line);
	}

	String[] readFlds() {
		String line = readLine();
		return line.split(" ");
	}

	int[] readNums() {
		String[] flds = readFlds();
		int[] nums = new int[flds.length];
		for (int i = 0; i < flds.length; i++) {
			nums[i] = pint(flds[i]);
		}
		return nums;
	}

	void p(char c) {
		_out.print(c);
	}

	void pln(char c) {
		_out.println(c);
	}

	void p(double d) {
		_out.print(d);
	}

	void pln(double d) {
		_out.println(d);
	}

	void p(long l) {
		_out.print(l);
	}

	void pln(long l) {
		_out.println(l);
	}

	void p(String s) {
		_out.print(s);
	}

	void pln(String s) {
		_out.println(s);
	}

	static BufferedReader _in;
	static PrintWriter _out;

	public static void main(String[] args) {
		long start = System.currentTimeMillis();
		_in = new BufferedReader(new InputStreamReader(System.in));
		_out = new PrintWriter(System.out);
		new Main().solve();
		_out.flush();
		long end = System.currentTimeMillis();
		if (bElapsed) {
			System.err.println((end - start) + "ms");
		}
	}
}