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>
#include <stdlib.h>

typedef struct _Numbers {
	int size;
	int *s;
}Numbers;

int main(void) {
	int n, m, sw;
	Numbers Num;
	scanf("%d", &n);
	Num.s = (int *)malloc(sizeof(int)*n);
	scanf("%d", Num.s);
	Num.size = 1;
	for (int i = 1; i < n; i++) {
		sw = 1;
		scanf("%d", &m);
		for (int j = 0; j < Num.size; j++) {
			if (m == Num.s[j]) {
				sw = 0;
				break;
			}
		}
		if (sw) {
			Num.s[Num.size] = m;
			Num.size++;
		}
	}
	printf("%d\n", Num.size);
	return 0;
}