description: On an endless checkered sheet of paper, n cells are chosen and colored in three colors, where n is divisible by 3. It turns out that there are exactly n/3 marked cells of each of three colors! 
  
  Find the largest such k that it's possible to choose k/3 cells of each color, remove all other marked cells, and then select three rectangles with sides parallel to the grid lines so that the following conditions hold:
  
    * No two rectangles can intersect (but they can share a part of the boundary). In other words, the area of intersection of any two of these rectangles must be 0.
    * The i-th rectangle contains all the chosen cells of the i-th color and no chosen cells of other colors, for i = 1, 2, 3. 
  
  Input
  
  The first line of the input contains a single integer n — the number of the marked cells (3 ≤ n ≤ 10^5, n is divisible by 3).
  
  The i-th of the following n lines contains three integers x_i, y_i, c_i (|x_i|,|y_i| ≤ 10^9; 1 ≤ c_i ≤ 3), where (x_i, y_i) are the coordinates of the i-th marked cell and c_i is its color.
  
  It's guaranteed that all cells (x_i, y_i) in the input are distinct, and that there are exactly n/3 cells of each color.
  
  Output
  
  Output a single integer k — the largest number of cells you can leave.
  
  Examples
  
  Input
  
  
  9
  2 3 1
  4 1 2
  2 1 3
  3 4 1
  5 3 2
  4 4 3
  2 4 1
  5 2 2
  3 5 3
  
  
  Output
  
  
  6
  
  
  Input
  
  
  3
  1 1 1
  2 2 2
  3 3 3
  
  
  Output
  
  
  3
  
  Note
  
  In the first sample, it's possible to leave 6 cells with indexes 1, 5, 6, 7, 8, 9.
  
  In the second sample, it's possible to leave 3 cells with indexes 1, 2, 3.
solution: #include <bits/stdc++.h>
  #pragma GCC optimize("O3")
  #pragma GCC target("sse4")
  using namespace std;
  template <class T>
  using pq = priority_queue<T>;
  template <class T>
  using pqg = priority_queue<T, vector<T>, greater<T>>;
  template <class T>
  bool ckmin(T& a, const T& b) {
    return b < a ? a = b, 1 : 0;
  }
  template <class T>
  bool ckmax(T& a, const T& b) {
    return a < b ? a = b, 1 : 0;
  }
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  void __print(int x) { cerr << x; }
  void __print(long x) { cerr << x; }
  void __print(long long x) { cerr << x; }
  void __print(unsigned x) { cerr << x; }
  void __print(unsigned long x) { cerr << x; }
  void __print(unsigned long long x) { cerr << x; }
  void __print(float x) { cerr << x; }
  void __print(double x) { cerr << x; }
  void __print(long double x) { cerr << x; }
  void __print(char x) { cerr << '\'' << x << '\''; }
  void __print(const char* x) { cerr << '\"' << x << '\"'; }
  void __print(const string& x) { cerr << '\"' << x << '\"'; }
  void __print(bool x) { cerr << (x ? "true" : "false"); }
  template <typename T, typename V>
  void __print(const pair<T, V>& x) {
    cerr << '{';
    __print(x.first);
    cerr << ", ";
    __print(x.second);
    cerr << '}';
  }
  template <typename T>
  void __print(const T& x) {
    int first = 0;
    cerr << '{';
    for (auto& i : x) cerr << (first++ ? ", " : ""), __print(i);
    cerr << "}";
  }
  void _print() { cerr << "]\n"; }
  template <typename T, typename... V>
  void _print(T t, V... v) {
    __print(t);
    if (sizeof...(v)) cerr << ", ";
    _print(v...);
  }
  const int MOD = 1000000007;
  const char nl = '\n';
  const int MX = 100001;
  int N;
  vector<int> A[2][3][2];
  bool ok(int K) {
    for (int i = 0; i < (2); i++) {
      for (int a = 0; a < (3); a++) {
        for (int b = 0; b < (3); b++) {
          if (a == b) continue;
          int c = 3 - a - b;
          int lstA = A[i][a][i][K - 1];
          int pB = upper_bound(A[i][b][i].begin(), A[i][b][i].end(), lstA) -
                   A[i][b][i].begin();
          pB += K - 1;
          if (pB < N / 3) {
            int lstB = A[i][b][i][pB];
            int pC = upper_bound(A[i][c][i].begin(), A[i][c][i].end(), lstB) -
                     A[i][c][i].begin();
            pC += K - 1;
            if (pC < N / 3) {
              return true;
            }
          }
          vector<int> okA, okB;
          int capOff = A[1 - i][c][1 - i][K - 1];
          for (int x = 0; x < (N / 3); x++) {
            if (A[i][a][1 - i][x] > capOff) {
              okA.push_back(A[i][a][i][x]);
            }
            if (A[i][b][1 - i][x] > capOff) {
              okB.push_back(A[i][b][i][x]);
            }
          }
          if ((int)(okA).size() >= K && (int)(okB).size() >= K &&
              okA[K - 1] < okB[(int)(okB).size() - K]) {
            return true;
          }
          okA.clear();
          okB.clear();
          capOff = A[1 - i][c][1 - i][N / 3 - K];
          for (int x = 0; x < (N / 3); x++) {
            if (A[i][a][1 - i][x] < capOff) {
              okA.push_back(A[i][a][i][x]);
            }
            if (A[i][b][1 - i][x] < capOff) {
              okB.push_back(A[i][b][i][x]);
            }
          }
          if ((int)(okA).size() >= K && (int)(okB).size() >= K &&
              okA[K - 1] < okB[(int)(okB).size() - K]) {
            return true;
          }
        }
      }
    }
    return false;
  }
  void solve() {
    cin >> N;
    int X[N], Y[N], C[N];
    for (int i = 0; i < (N); i++) cin >> X[i] >> Y[i] >> C[i];
    set<int> vX, vY;
    for (int i = 0; i < (N); i++) {
      vX.insert(X[i]), vY.insert(Y[i]);
    }
    map<int, int> cX, cY;
    for (auto& a : vX) {
      cX[a] = (int)(cX).size() + 1;
    }
    for (auto& a : vY) {
      cY[a] = (int)(cY).size() + 1;
    }
    vector<pair<int, int>> xSor[3], ySor[3];
    for (int i = 0; i < (N); i++) {
      pair<int, int> cur = {cX[X[i]], cY[Y[i]]};
      xSor[C[i] - 1].push_back({cur.first, cur.second});
      ySor[C[i] - 1].push_back({cur.second, cur.first});
    }
    for (int i = 0; i < (3); i++) {
      sort(xSor[i].begin(), xSor[i].end());
      sort(ySor[i].begin(), ySor[i].end());
      for (int j = 0; j < (N / 3); j++) {
        A[0][i][0].push_back(xSor[i][j].first);
        A[0][i][1].push_back(xSor[i][j].second);
        A[1][i][0].push_back(ySor[i][j].second);
        A[1][i][1].push_back(ySor[i][j].first);
      }
    }
    int lo = 1, hi = N / 3;
    while (lo < hi) {
      int mid = (lo + hi + 1) / 2;
      if (ok(mid)) {
        lo = mid;
      } else
        hi = mid - 1;
    }
    cout << lo * 3 << nl;
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int T = 1;
    while (T--) {
      solve();
    }
    return 0;
  }
  