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;
	int d[100];
	
	scanf("%d",&n);
	for(i=0;i<n;i++){
		scanf("%d",&d[i]);
	}
	d[n]=1000;
	
	for(i=0;i<n;i++){
		for(j=0;j<n-i;j++){
			if(d[j]>d[j+1]){
				tmp=d[j];
				d[j]=d[j+1];
				d[j+1]=tmp;
			}
		}
	}
	
	tmp=0;
	j=0;
	for(i=0;i<n;i++){
		if(d[i]>tmp){
			j++;
			tmp=d[i];
		}
	}
	
	printf("%d\n",j);
	
	return 0;
}