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>
  int n, ans;
  struct node {
    int x, y, c, d;
  } a[100005];
  bool solve1(int lim) {
    int l = 0, c = 0;
    while (l != n && c != lim) {
      if (a[l].c == 0) ++c;
      ++l;
    }
    while (l && l != n && a[l].x == a[l - 1].x) ++l;
    if (c != lim) return false;
    std::vector<node> t(a + l, a + n);
    std::sort(t.begin(), t.end(),
              [&](const node &A, const node &B) { return A.y < B.y; });
    l = 0;
    c = 0;
    while (l != t.size() && c != lim) {
      if (t[l].c == 1) ++c;
      ++l;
    }
    while (l && l != t.size() && t[l].y == t[l - 1].y) ++l;
    if (c != lim) return false;
    c = 0;
    while (l != t.size()) {
      if (t[l].c == 2) ++c;
      ++l;
    }
    return c >= lim;
  }
  bool solve2(int lim) {
    int l = 0, c = 0;
    while (l != n && c != lim) {
      if (a[l].c == 0) ++c;
      ++l;
    }
    while (l && l != n && a[l].x == a[l - 1].x) ++l;
    if (c != lim) return false;
    c = 0;
    while (l != n && c != lim) {
      if (a[l].c == 1) ++c;
      ++l;
    }
    while (l && l != n && a[l].x == a[l - 1].x) ++l;
    if (c != lim) return false;
    c = 0;
    while (l != n && c != lim) {
      if (a[l].c == 2) ++c;
      ++l;
    }
    return c == lim;
  }
  bool check(int x) {
    for (int i = 0; i != 4; ++i) {
      for (int j = 0; j != n; ++j) std::swap(a[j].x, a[j].y), a[j].x *= -1;
      std::sort(a, a + n,
                [&](const node &A, const node &B) { return A.x < B.x; });
      int p[3] = {0, 1, 2};
      do {
        for (int k = 0; k != n; ++k) a[k].c = p[a[k].d];
        if (solve1(x) || solve2(x)) return true;
      } while (std::next_permutation(p, p + 3));
    }
    return false;
  }
  signed main() {
    std::ios::sync_with_stdio(false);
    std::cin >> n;
    for (int i = 0; i != n; ++i) std::cin >> a[i].x >> a[i].y >> a[i].d, --a[i].d;
    int l = 1, r = n / 3;
    while (l < r) {
      int mid = (l + r + 1) >> 1;
      if (check(mid))
        l = mid;
      else
        r = mid - 1;
    }
    std::cout << l * 3 << std::endl;
    return 0;
  }
  