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: Accepted)

#include <stdio.h>
#include <stdlib.h>
int num(){
    char s[11];
    int i, n;
    for(i=0; i<sizeof(s); i++){
        s[i] = getchar();
        if(s[i] == ' ' || s[i] == '\n'){
            s[i] = '\0';
            break;
        }
    }
    sscanf(s, "%d", &n);
    return n;
}
void sort(int x[], int n){
    int y, i, j;
    for(i=n; i>1; i--){
        for(j=0; j<=i-2; j++){
            if(x[j] < x[j+1]){
                y = x[j];
                x[j] = x[j+1];
                x[j+1] = y;
            }
        }
    }
}
int main(void){
    int n;
    int d[101];
    int i, cou=1;
    
    n  = num();
    for(i=0; i<n; i++){
        d[i] = num();
    }
    
    sort(d, n);
    
    for(i=1; i<n; i++){
        if(d[i-1] > d[i]){
            cou++;
        }
    }
    
    printf("%d\n", cou);
    
    return 0;
}