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: Wrong Answer)

#include <stdio.h>

int main(void)
{
    int n;
    int i[100] = {0};

    scanf("%d", &n);
    int cnt;
    for (cnt = 0;cnt < n;cnt++) {
        scanf("%d", &i[cnt]);
    }

    int cnt2;
    int tmp;
    for (cnt = 0;cnt < n;cnt++) {
        for (cnt2 = cnt + 1;cnt2 < n;cnt2++) {
            if (i[cnt] > i[cnt2]) {
                tmp = i[cnt];
                i[cnt] = i[cnt2];
                i[cnt2] = tmp;
            }
        }
    }

    int ans = 1;
    for (cnt = 0;cnt < n;cnt++) {
        if ((i[cnt] - i[cnt + 1]) <= -2) {
            ans++;
        }
    }
    printf("%d\n", ans);
    return 0;
}