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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
class Program
{
    static void Main(string[] args)
    {
            // 入力
            string sS = Console.ReadLine().Trim();
            int iN = int.Parse(sS);
            List<int> lstInp = new List<int>();
            for (int i = 0; i < iN; i++)
            {
                sS = Console.ReadLine().Trim();
                lstInp.Add(int.Parse(sS));
            }

            System.Diagnostics.Stopwatch sw = new System.Diagnostics.Stopwatch();
            sw.Start();

            // 解析
            lstInp.Sort(delegate (int mca1, int mca2) { return (int)(mca2 - mca1); });      // 降順
            int iOut = 1;
            int iBak = lstInp[0];                                                           // 比較元
            for (int i = 1; i < lstInp.Count; i++)
            {
                if (iBak <= lstInp[i]) { continue; }                                        // 同じ大きさ以上はスルー
                iOut++;
                iBak = lstInp[i];
            }

            // 出力
            Console.WriteLine("{0}", iOut.ToString().Trim());

            sw.Stop();
#if DEBUG
            Console.WriteLine();
            Console.WriteLine("{0}", sw.Elapsed.ToString());
            Console.WriteLine("続行するには何かキーを押してください．．．");
            Console.ReadKey();
#endif
    }
}