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;
  template <typename T>
  struct fenwick {
    vector<vector<T>> a;
    int n;
    fenwick(int n) : n(n), a(2, vector<T>(n)) {}
    void add(int x, int p, T v) {
      assert(0 <= x && x < n);
      for (int i = x + 1; i <= n; i += i & -i) a[p][i - 1] += v;
    }
    T qry(int p, int x) {
      T ret = 0;
      for (int i = min(x + 1, n); i > 0; i -= i & -i) ret += a[p][i - 1];
      return ret;
    }
    int qry2() {
      int lg = 31 - __builtin_clz(n), pos = 0, s1 = qry(1, n - 1), ls = 0, rs = 0;
      for (int i = 1 << lg; i; i >>= 1) {
        if (pos + i <= n &&
            ls + a[0][pos + i - 1] <= s1 - rs - a[1][pos + i - 1]) {
          ls += a[0][pos + i - 1];
          rs += a[1][pos + i - 1];
          pos += i;
        }
      }
      int ret = min(ls, s1 - rs);
      if (pos < n) {
        ret = max(ret, min(qry(0, pos), s1 - qry(1, pos)));
      }
      return ret;
    }
  };
  struct node {
    int x, y, c;
    bool operator<(const node& n) const { return x < n.x; }
  };
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int n;
    cin >> n;
    vector<node> a(n);
    vector<int> b;
    for (int i = 0; i < n; ++i) {
      cin >> a[i].x >> a[i].y >> a[i].c;
      a[i].c -= 1;
      b.push_back(a[i].x);
      b.push_back(a[i].y);
    }
    sort(b.begin(), b.end());
    b.resize(unique(b.begin(), b.end()) - b.begin());
    for (int i = 0; i < n; ++i) {
      a[i].x = lower_bound(b.begin(), b.end(), a[i].x) - b.begin();
      a[i].y = lower_bound(b.begin(), b.end(), a[i].y) - b.begin();
    }
    int m = b.size(), ret = 0;
    for (int qaq = 0; qaq < 2; ++qaq) {
      sort(a.begin(), a.end());
      for (int tat = 0; tat < 2; ++tat) {
        for (int c1 = 0; c1 < 3; ++c1) {
          int c2 = (c1 + 1) % 3, c3 = (c2 + 1) % 3;
          for (int qwq = 0; qwq < 2; ++qwq, swap(c2, c3)) {
            assert(c1 != c2 && c2 != c3 && c1 != c3);
            fenwick<int> fx(m), fy(m);
            for (int i = 0; i < n; ++i) {
              if (a[i].c != c1) {
                fx.add(a[i].x, a[i].c == c2, 1);
                fy.add(a[i].y, a[i].c == c2, 1);
              }
            }
            int cnt1 = 0;
            for (int i = 0; i < n;) {
              int j = i;
              while (j < n && a[i].x == a[j].x) {
                if (a[j].c != c1) {
                  fx.add(a[j].x, a[j].c == c2, -1);
                  fy.add(a[j].y, a[j].c == c2, -1);
                }
                cnt1 += a[j].c == c1;
                ++j;
              }
              ret = max(ret, min(cnt1, fx.qry2()));
              ret = max(ret, min(cnt1, fy.qry2()));
              i = j;
            }
          }
        }
        reverse(a.begin(), a.end());
      }
      for (int i = 0; i < n; ++i) swap(a[i].x, a[i].y);
    }
    cout << ret * 3 << '\n';
  }
  