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;
  const int MAXN = (int)1e5 + 5;
  struct Point {
    int x, y, c, idx;
    Point() { x = y = c = idx = 0; }
    void rotate() {
      swap(x, y);
      x = -x;
    }
  } P[MAXN], Q[MAXN];
  bool operator<(Point a, Point b) {
    return make_pair(a.x, a.y) < make_pair(b.x, b.y);
  }
  bool byY(Point a, Point b) { return make_pair(a.y, a.x) < make_pair(b.y, b.x); }
  int perm[3];
  int n;
  bool check(int k) {
    iota(perm, perm + 3, 1);
    do {
      int p = 0, cnt = 0;
      while (p < n && (cnt < k || P[p].x == P[p + 1].x)) {
        p++;
        cnt += (P[p].c == perm[0]);
      }
      int q = p;
      cnt = 0;
      while (q < n && (cnt < k || P[q].x == P[q + 1].x)) {
        q++;
        cnt += (P[q].c == perm[1]);
      }
      cnt = 0;
      for (int i = q + 1; i <= n; i++) {
        cnt += (P[i].c == perm[2]);
      }
      if (cnt >= k) {
        return 1;
      }
      vector<Point> T;
      for (int i = 1; i <= n; i++) {
        if (Q[i].idx > p) {
          T.push_back(Q[i]);
        }
      }
      q = -1;
      cnt = 0;
      while (q + 1 < (int)(T).size() && (cnt < k || T[q].y == T[q + 1].y)) {
        q++;
        cnt += (T[q].c == perm[1]);
      }
      cnt = 0;
      for (int i = q + 1; i < (int)(T).size(); i++) {
        cnt += (T[i].c == perm[2]);
      }
      if (cnt >= k) {
        return 1;
      }
    } while (next_permutation(perm, perm + 3));
    return 0;
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cin >> n;
    for (int i = 1; i <= n; i++) {
      cin >> P[i].x >> P[i].y >> P[i].c;
    }
    int ans = 0;
    for (int i = 0; i < 3; i++) {
      perm[i] = i;
    }
    for (int rot = 0; rot < 4; rot++) {
      sort(P + 1, P + n + 1);
      for (int i = 1; i <= n; i++) {
        P[i].idx = i;
        Q[i] = P[i];
      }
      sort(Q + 1, Q + n + 1, byY);
      int l = 1, r = n / 3, opt = 0;
      while (l <= r) {
        int mid = (l + r) / 2;
        if (check(mid)) {
          opt = mid;
          l = mid + 1;
        } else {
          r = mid - 1;
        }
      }
      ans = max(ans, opt);
      for (int i = 1; i <= n; i++) {
        P[i].rotate();
      }
    }
    cout << ans * 3 << endl;
    return 0;
  }
  