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;
	int s[150];
	int s2[159];
	int same=0;
	int result=0;
	int min;
	scanf("%d\n",&n);
	for(i=0;i<n;i++)
	{
		scanf("%d\n",&s[i]);
	}
	for(i=0;i<=100;i++)
	{
		s2[i]=0;
	}

	
	for(i=0;i<n;i++)
	{
	for(j=1;j<=100;j++)
	{
		if(s[i]==j){
			s2[j]++;
	}
	}
	}
	result=0;
	for(i=1;i<=100;i++)
	{
		if(s2[i]!=0){
			result++;}
	}
	printf("%d",result);
	return 0;
	}