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;
  struct segtree {
    struct node {
      int s1 = 0, s2 = 0;
      void upd(int l, int r, int v1, int v2) {
        s1 += v1;
        s2 += v2;
      }
    };
    node merge(const node &x, const node &y) {
      node ret;
      ret.s1 = x.s1 + y.s1;
      ret.s2 = x.s2 + y.s2;
      return ret;
    }
    void build(int l, int r, int rt) {
      if (l == r) {
        return;
      }
      int m = (l + r) / 2;
      build(l, m, rt << 1);
      build(m + 1, r, rt << 1 | 1);
      a[rt] = merge(a[rt << 1], a[rt << 1 | 1]);
    }
    template <typename T>
    void build(int l, int r, int rt, const vector<T> &v) {
      if (l == r) {
        return;
      }
      int m = (l + r) / 2;
      build(l, m, rt << 1, v);
      build(m + 1, r, rt << 1 | 1, v);
      a[rt] = merge(a[rt << 1], a[rt << 1 | 1]);
    }
    segtree(int n) : n(n), a(n << 2) { build(0, n - 1, 1); }
    template <typename T>
    segtree(const vector<T> &v) : n(v.size()), a(v.size() << 2) {
      build(0, n - 1, 1, v);
    }
    int n;
    vector<node> a;
    template <typename... V>
    void update(int L, int R, int l, int r, int rt, const V &...v) {
      if (L <= l && r <= R) {
        a[rt].upd(l, r, v...);
        return;
      }
      int m = (l + r) / 2;
      if (L <= m) update(L, R, l, m, rt << 1, v...);
      if (m < R) update(L, R, m + 1, r, rt << 1 | 1, v...);
      a[rt] = merge(a[rt << 1], a[rt << 1 | 1]);
    }
    template <typename... V>
    void upd(int L, int R, const V &...v) {
      update(L, R, 0, n - 1, 1, v...);
    }
    int query(int l, int r, int rt, int sl, int sr) {
      if (l == r) return max(min(a[rt].s1 + sl, sr), min(sl, a[rt].s2 + sr));
      node ret;
      int m = (l + r) / 2;
      if (sl + a[rt << 1].s1 < sr + a[rt << 1 | 1].s2)
        return query(m + 1, r, rt << 1 | 1, sl + a[rt << 1].s1, sr);
      return query(l, m, rt << 1, sl, sr + a[rt << 1 | 1].s2);
    }
    int query2(int l, int r, int rt, int sl, int sr) {
      if (l == r) return max(min(a[rt].s2 + sl, sr), min(sl, a[rt].s1 + sr));
      node ret;
      int m = (l + r) / 2;
      if (sl + a[rt << 1].s2 < sr + a[rt << 1 | 1].s1)
        return query2(m + 1, r, rt << 1 | 1, sl + a[rt << 1].s2, sr);
      return query2(l, m, rt << 1, sl, sr + a[rt << 1 | 1].s1);
    }
    int qry() { return max(query(0, n - 1, 1, 0, 0), query2(0, n - 1, 1, 0, 0)); }
  };
  struct node {
    int x, y, c;
    bool operator<(const node &n) const { return x < n.x; }
    friend ostream &operator<<(ostream &os, node n) {
      os << "(" << n.x << "," << n.y << "," << n.c << ")";
      return os;
    }
  };
  int solve1(vector<node> a, int m) {
    int ret = 0, n = a.size();
    sort(a.begin(), a.end());
    for (int qwq = 0; qwq < 2; ++qwq) {
      for (int c1 = 1; c1 <= 3; ++c1) {
        int c2 = c1 % 3 + 1, c3 = c2 % 3 + 1;
        segtree st(m);
        for (auto [x, y, z] : a) {
          st.upd(y, y, z == c2, z == c3);
        }
        int cnt1 = 0;
        for (int i = 0, j; i < n; i = j) {
          j = i;
          while (j < n && a[j].x == a[i].x) {
            if (a[j].c == c1) {
              cnt1++;
            } else {
              st.upd(a[j].y, a[j].y, -(a[j].c == c2), -(a[j].c == c3));
            }
            ++j;
          }
          ret = max(ret, min(cnt1, st.qry()));
        }
      }
      reverse(a.begin(), a.end());
    }
    return ret;
  }
  int solve2(vector<node> a, int m) {
    int ret = 0, n = a.size();
    sort(a.begin(), a.end());
    for (int qwq = 0; qwq < 2; ++qwq) {
      for (int c1 = 1; c1 <= 3; ++c1) {
        int c2 = c1 % 3 + 1, c3 = c2 % 3 + 1;
        segtree st(m);
        for (auto [x, y, z] : a) {
          st.upd(x, x, z == c2, z == c3);
        }
        int cnt1 = 0;
        for (int i = 0, j; i < n; i = j) {
          j = i;
          while (j < n && a[j].x == a[i].x) {
            if (a[j].c == c1) {
              cnt1++;
            } else {
              st.upd(a[j].x, a[j].x, -(a[j].c == c2), -(a[j].c == c3));
            }
            ++j;
          }
          ret = max(ret, min(cnt1, st.qry()));
        }
      }
      reverse(a.begin(), a.end());
    }
    return ret;
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int n;
    cin >> n;
    vector<node> a(n);
    vector<int> xx, yy;
    for (auto &[x, y, z] : a) {
      cin >> x >> y >> z;
      xx.push_back(x);
      yy.push_back(y);
    }
    sort(xx.begin(), xx.end());
    xx.resize(unique(xx.begin(), xx.end()) - xx.begin());
    sort(yy.begin(), yy.end());
    yy.resize(unique(yy.begin(), yy.end()) - yy.begin());
    for (auto &[x, y, z] : a) {
      x = lower_bound(xx.begin(), xx.end(), x) - xx.begin();
      y = lower_bound(yy.begin(), yy.end(), y) - yy.begin();
    }
    sort(a.begin(), a.end());
    int ans = 0;
    ans = max(ans, solve1(a, yy.size()));
    ans = max(ans, solve2(a, xx.size()));
    for (auto &[x, y, z] : a) {
      swap(x, y);
    }
    ans = max(ans, solve1(a, xx.size()));
    ans = max(ans, solve2(a, yy.size()));
    cout << ans * 3 << '\n';
  }
  