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: Wrong Answer)

#include<iostream>
using namespace std;
#include<iostream>
using namespace std;
#define MAX 500000
#define SENTINEL 2000000000

int L[MAX/2 + 2], R[MAX/2+2];
int cnt;

void merge(int A[], int n, int left, int mid, int right){
  int n1 = mid - left;
  int n2 = right - mid;
  for( int i = 0; i < n1; i++ )L[i] = A[left + i];
  for( int i = 0; i < n2; i++ )R[i] = A[mid + i];
  L[n1] = R[n2] = SENTINEL;
  int i = 0, j = 0;
  for ( int k = left; k < right; k++ ){
    cnt++;
    if(L[i] <= R[j]){
      A[k] = L[i++];
    }else{
      A[k] = R[j++];
    }
  }
}

void mergeSort(int A[], int n, int left, int right){
  if( left+1 < right){
    int mid = (left + right) / 2;
    mergeSort(A, n, left, mid);
    mergeSort(A, n, mid, right);
    merge(A, n, left, mid, right);
  }
}

int main(){
  int A[MAX], n, i;
  int m=0;
  cin >> n;
  for ( i = 0; i < n; i++) cin >> A[i];
  mergeSort(A, n, 0, n);
  for(int i=0;i<n;i++){
    cin>>A[i];
    if(A[i]<A[i+1])m++;
  }
  cout<<m<<endl;
  return 0;
}