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 (Clang 10.0.0) (judge verdict: Accepted)

#include<stdio.h>
int main()
{
 int N,i,j,sum,tmp;
  int a[100]={0};
  scanf("%d",&N);
  sum=N;
  for(i=0;i<N;i++)
  {
    scanf("%d",&a[i]);
  }
  for(i=0;i<N-1;i++)
  {
    for(j=0;j<N-1;j++)
    {
     if(a[j]<=a[j+1])
     {
       tmp=a[j];
       a[j]=a[j+1];
       a[j+1]=tmp;
     } 
    }
  }
  for(i=0;i<N-1;i++)
  {
  if(a[i]==a[i+1])
  {
    sum=sum-1;
  }
  }
  printf("%d",sum);
  return 0;
}