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 9.2.1) (judge verdict: Accepted)

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

int main() {
	int N;
    cin >> N;
	vector<int> kagamimochi(N);
    for(int i = 0; i < N; i++){
    	cin >> kagamimochi.at(i);
    }
    int Max , Count;
    Count = 0;
    for(int i = 0; i < N; i++){
    	Max = kagamimochi.at(i);
        for(int j = 0; j < N; j++){
        	if(Max < kagamimochi.at(j)){
            	Max = kagamimochi.at(j);
            }
        }
        if(Max == 0){
        	break;
        }
        for(int k = 0; k < N; k++){
        	if(kagamimochi.at(k) == Max){
            	kagamimochi.at(k) = 0;
            }
        }
        Count++;
    }
  	cout << Count << endl;
}