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)

/*
strlenはcharへのポインタが引数なので、今回の
intの配列は渡せない
*/

#include<stdio.h>
#include<string.h>

void motiSort(int *p, int n);

int main(void)
{
  int n = 0;
  int wid[100] = {0};
  int i;//for loop
  int sum = 1;

  scanf("%d", &n);
  for(i=0; i<n; i++){
    scanf("%d", wid+i);
  }
  //printf("***\n");
  motiSort(wid, n);

  for(i=1; i<n; i++){
    if(wid[i-1] != wid[i]) sum++;
  }

  printf("%d\n", sum);
}

void motiSort(int *p , int n)
{
  int i = 1, buffer = 0;
  int j;

  while(i < n){
    j = i;
    while(j > 0){
      if(p[j-1] > p[j]){
        buffer = p[j-1];
        p[j-1] = p[j];
        p[j] = buffer;
      }
      j--;
    }
    i++;
  }
}