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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

int main() {
  int N = 0;
  cin >> N;
  int count = N;
  
  vector<int> d(N);
  for (int i = 0; i < N; i++) {
    cin >> d.at(i);
  }
  
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < N - 1; j++) {
      if (d.at(N - j - 1) > d.at(N - j - 2)) {
        int tmp = d.at(N - j - 2);
        d.at(N - j - 2) = d.at(N - j - 1);
        d.at(N - j - 1) = tmp;
      }
    }
  }
  
  for (int i = 0; i < N - 1; i++) {
    if (d.at(i) == d.at(i + 1)) {
      count--;
    }
  }
  
  cout << count << endl;
}