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 shellsort(int a[], int l, int r);

int main(void)
{
  char a[5];
  int n;
  int d[110]={0};

  fgets(a, sizeof(a), stdin);
  sscanf(a, "%d", &n);

  for (int i = 0; i < n; i++) {
    scanf("%d", &d[i]);
  }
  shellsort(d, 0, n-1);
  int v = 0;
  for (int i = 0; i < n; i++) {
    if (d[i] == v) {
      d[i] = 0;
      continue;
    }
    v = d[i];
  }
  shellsort(d, 0, n-1);
  int ans=0;
  for (int i = 0; d[i] != 0; i++) {
    ans++;
  }


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

    return 0;
}

void shellsort(int a[], int l, int r)
{
  int h;
  for (h=1; h <= (r-l)/9; h = 3*h+1)
      ;
      for (; h > 0; h /= 3) {
        for (int i = l+h; i <= r; i++) {
          int j = i;
          int v = a[i];
          while (j >= l+h && v > a[j-h]) {
            a[j] = a[j-h];
            j -= h;
          }
          a[j] = v;
        }
      }
}