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 <algorithm>
#include <bits/stdc++.h>
#include <climits>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <list>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <vector>
using namespace std;

#define rep(X, S, E) for (int(X) = (S); (X) < (E); ++(X))
#define rrep(X, S, E) for (int(X) = (E)-1; (X) >= (S); --(X))
#define arep(X, Y) for (auto(X) : Y)
#define all(X) (X).begin(), (X).end()
#define pb push_back
#define mp make_pair
#define fi first
#define sc second
#define print(x) cout << x << endl
#define printDouble(x) cout << fixed << setprecision(13) << x << endl
#define chmin(x, y) x = min(x, y)
#define chmax(x, y) x = max(x, y)

using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using vi = vector<int>;
using vl = vector<ll>;

using decendingQueue = priority_queue<ll, vl>;              //降順
using ascendingQueue = priority_queue<ll, vl, greater<ll>>; //昇順

const int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
const int INT_INF = 1e9;
const ll LL_INF = 1e18;
const int mod = 1000000007;

template <class T> void mySort(vector<T> &X, bool isAscending) {
  // 昇順
  if (isAscending) {
    sort(all(X));
  } else {
    // 降順
    sort(all(X), greater<T>());
  }
}

long long gcd(long long m, long long n) {
  if (m < n)
    return gcd(n, m);
  if (n == 0)
    return m;
  return gcd(n, m % n);
}

long long lcm(long long m, long long n) {
  // m * nでlong型のオーバフローを発生させないため、先に割り算から行う
  return m * (n / gcd(m, n));
}

// auto mod int
// https://youtu.be/L8grWxBlIZ4?t=9858
// https://youtu.be/ERZuLAxZffQ?t=4807 : optimize
// https://youtu.be/8uowVvQ_-Mo?t=1329 : division
struct mint {
  ll x; // typedef long long ll;
  mint(ll x = 0) : x((x % mod + mod) % mod) {}
  mint &operator+=(const mint a) {
    if ((x += a.x) >= mod)
      x -= mod;
    return *this;
  }
  mint &operator-=(const mint a) {
    if ((x += mod - a.x) >= mod)
      x -= mod;
    return *this;
  }
  mint &operator*=(const mint a) {
    (x *= a.x) %= mod;
    return *this;
  }
  mint operator+(const mint a) const {
    mint res(*this);
    return res += a;
  }
  mint operator-(const mint a) const {
    mint res(*this);
    return res -= a;
  }
  mint operator*(const mint a) const {
    mint res(*this);
    return res *= a;
  }
  mint pow(ll t) const {
    if (!t)
      return 1;
    mint a = pow(t >> 1);
    a *= a;
    if (t & 1)
      a *= *this;
    return a;
  }

  // for prime mod
  mint inv() const { return pow(mod - 2); }
  mint &operator/=(const mint a) { return (*this) *= a.inv(); }
  mint operator/(const mint a) const {
    mint res(*this);
    return res /= a;
  }
};

// combination mod prime
// https://www.youtube.com/watch?v=8uowVvQ_-Mo&feature=youtu.be&t=1619
struct combination {
  vector<mint> fact, ifact;
  combination(int n) : fact(n + 1), ifact(n + 1) {
    assert(n < mod);
    fact[0] = 1;
    for (int i = 1; i <= n; ++i)
      fact[i] = fact[i - 1] * i;
    ifact[n] = fact[n].inv();
    for (int i = n; i >= 1; --i)
      ifact[i - 1] = ifact[i] * i;
  }
  mint operator()(int n, int k) {
    if (k < 0 || k > n)
      return 0;
    return fact[n] * ifact[k] * ifact[n - k];
  }
};

vector<long long> SieveOfEratosthenes(int max) {
  vector<long long> sieve;
  vector<long long> primes;

  for (int i = 1; i < max + 1; ++i) {
    sieve.push_back(i);
  }

  sieve[0] = 0;
  for (int i = 2; i < max + 1; ++i) {
    if (sieve[i - 1] != 0) {
      primes.push_back(sieve[i - 1]);
      for (int j = 2 * sieve[i - 1]; j < max + 1; j += sieve[i - 1]) {
        sieve[j - 1] = 0;
      }
    }
  }
  return primes;
}

class UnionFindTree {
private:
  vector<int> par;
  vector<int> rnk;
  vector<int> siz;

public:
  UnionFindTree(int n) {
    par.assign(n, -1);
    rnk.assign(n, -1);
    siz.assign(n, -1);

    for (int i = 0; i < n; ++i) {
      par[i] = i;
      rnk[i] = 0;
      siz[i] = 1;
    }
  }

  int find(int x) {
    if (par[x] == x)
      return x;
    else
      return par[x] = find(par[x]);
  }

  bool same(int x, int y) { return find(x) == find(y); }

  void unite(int x, int y) {
    x = find(x);
    y = find(y);
    if (x == y)
      return;
    if (rnk[x] < rnk[y]) {
      par[x] = y;
      siz[y] += siz[x];
    } else {
      par[y] = x;
      siz[x] += siz[y];
      if (rnk[x] == rnk[y])
        ++rnk[x];
    }
  }

  int size(int x) {
    x = find(x);
    return siz[x];
  }
};

class Edge {
public:
  ll from;
  ll to;
  ll cost;

  Edge() {}
  Edge(ll from, ll to, ll cost) {
    this->from = from;
    this->to = to;
    this->cost = cost;
  }

  bool operator<(const Edge &edge) const {
    return cost < edge.cost; //昇順
  }
};

class Graph {
public:
  ll nodes; // ノード数
  vector<Edge> edges;

  Graph() {}
  Graph(ll nodes) { this->nodes = nodes; }
  void addEdge(ll from, ll to, ll cost) {
    this->edges.push_back(Edge(from, to, cost));
  }
};

// クラスカル法
// 連結グラフの最小全域木を求める
class Kruskal {
private:
  Graph graph;
  vector<Edge> MinimumSpanningTree;
  ll minimumCost;
  void searchMinimumSpanningTree() {
    UnionFindTree uf(graph.nodes);
    sort(all(graph.edges));
    for (auto edge : graph.edges) {
      if (!uf.same(edge.from, edge.to)) {
        uf.unite(edge.from, edge.to);
        MinimumSpanningTree.push_back(edge);
      }
    }
  }

public:
  Kruskal(Graph graph) { this->graph = graph; }
  ll getMinimumSpanningTreeCost() {
    searchMinimumSpanningTree();
    ll cost = 0;
    for (auto tr : MinimumSpanningTree) {
      cost += tr.cost;
    }
    return cost;
  }
};

// ダイクストラ法 O((E+V)logV)
// 最小経路問題を解くためのアルゴリズム。辺の重みに負数を含む場合は利用不可
// 無向グラフの場合はコメントアウト箇所をコメントイン
class Dijkstra {
private:
  Graph graph;
  map<ll, vector<Edge>> fromPaths;
  vl distances;
  vl srcs;

public:
  Dijkstra(Graph graph) {
    this->graph = graph;
    for (auto edge : graph.edges) {
      fromPaths[edge.from].push_back(edge);
      // fromPaths[edge->to].push_back(Edge(edge->to, edge->from, edge->cost));
    }
  }
  void searchMinimumPathFrom(ll src) {
    // 複数回呼ばれる度に計算する
    distances = vl(graph.nodes + 1, LL_INF);
    srcs = vl(graph.nodes + 1, LL_INF);

    // 距離が近い順番に捜査していく
    priority_queue<pll, vector<pll>, greater<pll>> pq;
    distances[src] = 0;
    srcs[src] = -1;

    pq.push(mp(0ll, src));
    while (!pq.empty()) {
      int u = pq.top().second;
      double uw = pq.top().first;
      pq.pop();

      if (distances[u] < uw) {
        continue;
      };

      for (auto edge : fromPaths[u]) {
        int v = edge.to;
        ll w = edge.cost;
        if (distances[v] > distances[u] + w) {
          distances[v] = distances[u] + w;
          srcs[v] = u;
          pq.push(mp(distances[v], v));
        }
      }
    }
  };
  ll getDistance(ll n) { return distances[n]; }
  ll getFrom(ll n) { return srcs[n]; }
};

// ベルマンフォード O(|V||E|)
// 負コストが含まれていても最短経路問題を解くためのアルゴリズム。閉路の検出も可能
// 有向グラフ
class BellmanFord {
private:
  Graph graph;
  // 閉路が含まれるかは個々のノードごとに管理する必要あり
  vector<bool> hasNegativeCycles;
  vector<ll> distances;
  vl srcs;

public:
  BellmanFord(Graph graph) {
    ll nodes = graph.nodes + 1;

    this->graph = graph;
    distances = vector<ll>(nodes, LL_INF);
    hasNegativeCycles = vector<bool>(nodes, false);
  }
  void searchMinimumPathFrom(ll src) {
    distances[src] = 0;

    for (ll i = 0; i < graph.nodes - 1; i++) {
      for (auto edge : graph.edges) {
        ll u = edge.from;
        ll v = edge.to;
        ll w = edge.cost;
        if (distances[u] != LL_INF) {
          chmin(distances[v], distances[u] + w);
        }
      }
    }

    for (auto edge : graph.edges) {
      ll u = edge.from;
      ll v = edge.to;
      ll w = edge.cost;
      if (distances[u] == LL_INF) {
        continue;
      }
      if (distances[u] + w < distances[v]) {
        hasNegativeCycles[v] = true;
      }
    }

    for (ll i = 0; i < graph.nodes; i++) {
      for (auto edge : graph.edges) {
        ll u = edge.from;
        ll v = edge.to;
        ll w = edge.cost;
        if (distances[u] == LL_INF) {
          continue;
        }
        chmin(distances[v], distances[u] + w);
        if (hasNegativeCycles[u] == true) {
          hasNegativeCycles[v] = true;
        }
      }
    }
  }
  ll getDistance(ll n) { return distances[n]; }
  bool hasNegativeCycle(ll n) { return hasNegativeCycles[n]; }
  ll getFrom(ll n) { return srcs[n]; }
};

// O(V^3) 有向グラフ
class WarshallFloyd {
private:
  Graph graph;
  int nodes = this->graph.nodes + 1;
  vector<vector<ll>> distances;

public:
  WarshallFloyd(Graph graph) {
    this->graph = graph;
    nodes = this->graph.nodes + 1;
    this->distances = vector<vector<ll>>(nodes, vector<ll>(nodes, LL_INF));
    for (auto edge : graph.edges) {
      int from = edge.from;
      int to = edge.to;
      int cost = edge.cost;
      distances[from][to] = cost;
    }
  }
  void searchMinimumPath() {
    for (int k = 0; k < nodes; k++) {
      for (int i = 0; i < nodes; i++) {
        for (int j = 0; j < nodes; j++) {
          if (distances[i][k] == LL_INF || distances[k][j] == LL_INF) {
            continue;
          }
          if (distances[i][k] + distances[k][j] < distances[i][j]) {
            distances[i][j] = distances[i][k] + distances[k][j];
          }
        }
      }
    }
  }
  ll getDistance(int from, int to) { return distances[from][to]; }
};

void solve(long long N, std::vector<long long> d) {
  set<ll> dset(all(d));
  print(dset.size());
}

int main() {
  long long N;
  scanf("%lld", &N);
  std::vector<long long> d(N);
  for (int i = 0; i < N; i++) {
    scanf("%lld", &d[i]);
  }
  solve(N, std::move(d));
  return 0;
}