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>

using namespace std;

typedef long long int ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define rep2(i,n) for(int i=1;i<=(int)(n);i++)
#define rep3(i,n) for(int i=0;i<=(int)(n);i++)
#define REP(i,m,n) for(int i=m;i<(int)(n);i++)
#define repll(i,n) for(ll i=0;i<(ll)(n);i++)
#define repll2(i,n) for(ll i=1;i<=(ll)(n);i++)
#define REPLL(i,m,n) for(ll i=m;i<(ll)(n);i++)
#define INF (int)1e9  //
#define LINF (ll)4e18  //
#define MOD (int)(1e9)+7
#define MP make_pair
#define YES(n) cout << ((n) ? "YES" : "NO"  ) << endl
#define Yes(n) cout << ((n) ? "Yes" : "No"  ) << endl
#define yes(n) cout << ((n) ? "yes" : "no"  ) << endl
#define PI 3.1415926535898

//------------------------------------------------------

int main() {
    int n;
    cin >> n;
    set<int> st;

    rep(i, n) {
        int d;
        cin >> d;
        st.insert(d);
    }

    cout << st.size() << endl;

    return 0;
}