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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#include <numeric>
using namespace std;

#define ll long long
#define Fi first
#define Se second
#define Pb push_back
#define vi vector<int>
#define vvi vector<vector<int>>
#define vvp vector<vector<pair<int,int>>>

const int INF = 1e9;
const int MOD = 1e9+7;

#define all(v) v.begin(),v.end()
#define output_s(a) printf("%s\n",(a).c_str())

#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define Rep(i,n,m) for(int i=(int)(n);i<(int)(m);i++)
#define RepS(i,n,m) for(int i=(int)(n);i<=(int)(m);i++)
#define RRepS(i,n,m) for(int i=(int)(n);i>=(int)(m);i++)
#define RepA(x) for(auto &item:(x))

void CINT(){}
template <class Head,class... Tail>
void CINT(Head&& head,Tail&&... tail){
    cin>>head;
    CINT(move(tail)...);
}    
#define CIN(...) int __VA_ARGS__;CINT(__VA_ARGS__)
#define LCIN(...) ll __VA_ARGS__;CINT(__VA_ARGS__)
#define SCIN(...) string __VA_ARGS__;CINT(__VA_ARGS__)

template<typename TypeInt> //最大公約数
TypeInt GCD(TypeInt a,TypeInt b){
    if(a%b==0) return b;
    else return GCD(b,a%b);
}
template<typename TypeInt> //最小公倍数
TypeInt LCM(TypeInt a,TypeInt b){
    return a*b/GCD(a,b);
}

//library----------------------------------------------


signed main(){
    CIN(N);
    vi v(N);
    rep(i,N)cin>>v[i];
    sort(all(v),greater<int>());
    int big = INF,count=0;
    rep(i,N){
        if(big>v[i]){
            big=v[i];
            count++;
        }
    }
    printf("%d\n",count);
    return 0;
}