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>
  using namespace std;
  int n;
  vector<pair<int, int> > B[3], A[8][3];
  inline bool chk(int x) {
    int p[3] = {0, 1, 2};
    do {
      for (int o = 0; o < 4; o++) {
        int u = A[o << 1][p[0]][x - 1].first + 1;
        int y = x, v = ((int)1e9 + 1);
        for (auto [i, j] : A[o << 1][p[1]])
          if (i >= u && !--y) {
            v = i + 1;
            break;
          }
        y = x;
        for (auto [i, j] : A[o << 1][p[2]])
          if (i >= v && !--y) return 1;
        y = x, v = ((int)1e9 + 1);
        for (auto [i, j] : A[o << 1 | 1][p[1]])
          if (i >= u && !--y) {
            v = j + 1;
            break;
          }
        y = x;
        for (auto [i, j] : A[o << 1 | 1][p[2]])
          if (i >= u && j >= v && !--y) return 1;
      }
    } while (next_permutation(p, p + 3));
    return 0;
  }
  int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
      int x, y, c;
      scanf("%d%d%d", &x, &y, &c);
      B[c - 1].emplace_back(x, y);
    }
    for (int oo = 0; oo < 2; oo++) {
      for (int o = 0; o < 2; o++) {
        for (int i = 0; i < 3; i++) {
          A[(oo << 2) | (o << 1)][i] = B[i];
          sort(A[(oo << 2) | (o << 1)][i].begin(),
               A[(oo << 2) | (o << 1)][i].end());
          A[(oo << 2) | (o << 1) | 1][i] = B[i];
          sort(A[(oo << 2) | (o << 1) | 1][i].begin(),
               A[(oo << 2) | (o << 1) | 1][i].end(),
               [](const pair<int, int>& u, const pair<int, int>& v) {
                 return u.second == v.second ? u.first < v.first
                                             : u.second < v.second;
               });
        }
        for (int i = 0; i < 3; i++)
          for (int j = 0; j < n / 3; j++) B[i][j].first *= -1;
      }
      for (int i = 0; i < 3; i++)
        for (int j = 0; j < n / 3; j++) swap(B[i][j].first, B[i][j].second);
    }
    int L = 1, R = n / 3, res = L;
    while (L <= R) {
      int mid = (L + R) >> 1;
      if (chk(mid))
        res = mid, L = mid + 1;
      else
        R = mid - 1;
    }
    printf("%d\n", 3 * res);
  }
  