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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <stdio.h>
int main(void){
int i,j;
int n;
int d[101];
int e[101];
int x=0;
int z=0;
int v;
int temp;
scanf("%d",&n);
for(i=0;i<n;i++){
	
	scanf("%d",&d[i]);

}
for(i=0;i<n;i++){
	for(j=n-1;j>i;j--){
		if(d[j]>d[j-1]){
			
			temp=d[j];
			d[j]=d[j-1];
			d[j-1]=temp;
			
			
			
			
			
			
			
			
			
		}
		
		
		
		
		
		
		
	}
	
	
	
}

for(i=0;i<n;i++){
	if((i!=0)&&(d[i]!=d[i+1])&&(d[i+1]!=0)){
		printf("%d %d\n",d[i],d[i+1]);
	x++;
	}
	
	
}
printf("%d\n",x+1);
	return 0;
}