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 3.8.0) (judge verdict: Wrong Answer)

#include "stdio.h"

int sort(int a[], int x){
  int box;

  for(int i = 0; i < x; i++){
    for(int j = 1; j < x; j++){
      if(a[i] < a[j]){
        box =a[i];
        a[i] = a[j];
        a[j] = box;
      }
    }
  }

  return 0;
}

int main(void){
   int N;
   int count = 1;
   int d[120];

   scanf("%d", &N);

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

   sort(d,N);

   for(int j = 0; j < N-1; j++){
     if(d[j] != d[j+1]){
       count++;
     }
   }

   printf("%d\n",count);

   return 0;
}