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<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<algorithm>
#include<queue>
#include<vector>
#include<map>
#define ll long long
#define min(a,b) (a<b?a:b)
#define max(a,b) (a>b?a:b)
ll read()
{
	ll tmp=0; char f=1,c=getchar();
	while(c<'0'||'9'<c){if(c=='-')f=-1; c=getchar();}
	while('0'<=c&&c<='9'){tmp=tmp*10+c-'0'; c=getchar();}
	return tmp*f;
}
using namespace std;
int cnt[110];
int n;
int main()
{
	n=read();
	for(int i=1;i<=n;i++)++cnt[read()];
	int ans=0;
	for(int i=1;i<=100;i++)if(cnt[i])++ans;
	printf("%d\n",ans);
	return 0;
}