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>

int max(int N, int d[]);

int main(void){
  int N;
  int d[100] = {};
  int d_new[100] = {};//d[100]を大きい順に並べ替えたもの
  int i, j, k;
  int count = 0;

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

  j = max(N, d);
  d_new[0] = d[j];
  d[j] = 0;
  count++;

  for(i = 1; i < N; i++){
    do{
      j = max(N, d);
      d_new[i] = d[j];
      d[j] = 0;
    }while(d_new[i] == d_new[i-1]);
    if(d_new[i] != 0){
      count++;
    }else{
      break;
    }
  }

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

  return 0;
}

//d[i]を最大にするiのうち最小のものを求める関数
int max(int N, int d[]){
  int max;
  int i, j;

  max = d[0];
  for(i = 1; i < N; i++){
    if(d[i] > max){
      max = d[i];
    }
  }

  for(i = 0; i < N; i++){
    if(d[i] == max){
      j = i;
      break;
    }
  }

  return j;
}