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: Wrong Answer)

#include <stdio.h>

void swap_s(int *a, int *b){
  int temp = *a;
  *a = *b;
  *b = temp;
}

void sort_s(int *s, int n){
  int i, j;
  
  for(i = 0; i < n - 1; i++){
    for(j = n - i - 1; j > 0; j--){
      if(s[j - 1] > s[j])
        swap_s(&s[j - 1], &s[j]);
    }
  }
}

int main(){
  int n;
  int i;
  int a[128];
  int k = 0;
  
  scanf("%d", &n);
  
  for(i = 0; i < n; i++){
    scanf("%d", &a[i]);
  }
  
  sort_s(a, n);
  
  for(i = n - 1; i >= 0; i--){
    if(a[i - 1] < a[i])
      k++;
  }
  
  printf("%d\n", k);
  return 0;
}