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 main(){
  int n,i,j,tmp,now,count=0;
  int mochi[100];
  
  scanf("%d",&n);
  for(i=0;i<n;i++){
    scanf("%d",&mochi[i]);
  }
  
  for(i=n-1;i>0;i--){
    for(j=0;j<i;j++){
      if(mochi[j]<mochi[j+1]){
        tmp=mochi[j];
        mochi[j]=mochi[j+1];
        mochi[j+1]=tmp;
      }
    }
  }
  
  for(i=0;i<n;i++){
    if(i==0){
      now=mochi[i];
      count++;
    }
    
    if(now>mochi[i]){
      now=mochi[i];
      count++;
    }
  }
  printf("%d",count);
}