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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>

void quicksort(int number[],int first,int last)
{
   int i, j, pivot, temp;

   if(first < last){
      pivot = first;
      i = first;
      j = last;

      while(i < j){
         while((number[pivot] >= number[i]) && (i < last))
         {
            i++;
         }
         while(number[pivot] < number[j])
         {
            j--;
         }
         if(i < j){
            temp = number[i];
            number[i] = number[j];
            number[j] = temp;
         }
      }

      temp = number[pivot];
      number[pivot] = number[j];
      number[j] = temp;
      quicksort(number,first,j - 1);
      quicksort(number,j+1,last);

   }
}


int main()
{
    int n,d[100];

    scanf("%d",&n);
    for (int i = 0; i < n; i++)
    {
        scanf("%d",&d[i]);
    }
    quicksort(d,0,n - 1);

    int ans = n;
    for (int i = 0; i < n - 1; i++)
    {
        if (d[n - i - 1] == d[n - i - 2])
        {
            ans --;
        }
    }
    printf("%d",ans);
}