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... Args>
  void pr(Args... args) {}
  using ll = long long;
  const char nl = '\n';
  const int MX = 2e5 + 10;
  const int INF = 1e9 + 10;
  int n;
  array<int, 3> a[MX];
  void flip() {
    for (int i = 0; i < n; i++) swap(a[i][0], a[i][1]);
  }
  void flip2() {
    for (int i = 0; i < n; i++) a[i][0] = INF - a[i][0];
  }
  bool works(int k) {
    int one = n / 3;
    priority_queue<int> three;
    priority_queue<int, vector<int>, greater<int>> two;
    for (int i = n - 1, j = n - 1; i >= 0; i = j) {
      while (j >= 0 && a[i][0] == a[j][0]) {
        if (a[j][2] == 0) --one;
        if (a[j][2] == 1) two.push(a[j][1]);
        if (a[j][2] == 2) three.push(a[j][1]);
        if (((int)(two.size())) > k) two.pop();
        if (((int)(three.size())) > k) three.pop();
        --j;
      }
      if (one >= k && ((int)(two.size())) == k && ((int)(three.size())) == k &&
          two.top() > three.top()) {
        return true;
      }
    }
    return false;
  }
  int pre[MX], suf[MX];
  bool works2(int k) {
    fill_n(pre, n + 1, 0);
    fill_n(suf, n + 1, 0);
    for (int i = 0; i < n; i++) {
      if (a[i][2] == 0) pre[i + 1]++;
      if (a[i][2] == 2) suf[i]++;
    }
    for (int i = 0; i < n; i++) {
      pre[i + 1] += pre[i];
    }
    for (int i = n - 1; i >= 0; --i) {
      suf[i] += suf[i + 1];
    }
    int cur = 0;
    for (int i = 0, j = 0, l = 0, r = 0; i < n; i++) {
      while (j < n && cur < k) {
        cur += a[j][2] == 1;
        j++;
      }
      while (a[l][0] < a[i][0]) l++;
      while (r < n && a[r][0] <= a[j - 1][0]) r++;
      if (cur == k && pre[l] >= k && suf[r] >= k) {
        return true;
      }
      if (a[i][2] == 1) --cur;
    }
    return false;
  }
  void solve() {
    cin >> n;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < 3; j++) {
        cin >> a[i][j];
      }
      --a[i][2];
    }
    int ans = 0;
    for (int rep = 0; rep < 2; rep++) {
      for (int rep2 = 0; rep2 < 2; rep2++) {
        sort(a, a + n);
        vector<int> perm(3), iperm(3);
        iota(perm.begin(), perm.end(), 0);
        do {
          for (int i = 0; i < 3; i++) {
            iperm[perm[i]] = i;
          }
          for (int i = 0; i < n; i++) {
            a[i][2] = perm[a[i][2]];
          }
          int lo = 0, hi = n / 3 + 1;
          while (hi - lo > 1) {
            int mid = (lo + hi) / 2;
            if (works(mid) || works2(mid)) {
              lo = mid;
            } else {
              hi = mid;
            }
          }
          ans = max(ans, lo);
          for (int i = 0; i < n; i++) {
            a[i][2] = iperm[a[i][2]];
          }
        } while (next_permutation(perm.begin(), perm.end()));
        flip2();
      }
      flip();
    }
    cout << ans * 3 << nl;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int te = 1;
    while (te--) {
      solve();
    }
  }
  