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;
  long long a, b, q, w, e, ann, u[1200001], cn, fl, v1[1200001], v2[1200001];
  struct p {
    long long q, w, id;
  };
  vector<p> qu[4];
  struct seg {
    struct pp {
      long long sum;
    } l[2400001];
    void pushup(long long qq) { l[qq].sum = l[qq << 1].sum + l[qq << 1 | 1].sum; }
    void change(long long x, long long ll, long long rr, long long qq,
                long long ww) {
      if (ll == rr) {
        l[x].sum = ww;
        return;
      }
      long long mid = ((ll + rr) >> 1);
      if (mid >= qq)
        change(x << 1, ll, mid, qq, ww);
      else
        change(x << 1 | 1, mid + 1, rr, qq, ww);
      pushup(x);
    }
    long long query(long long x, long long ll, long long rr, long long qq) {
      if (l[x].sum < qq) return -1;
      if (ll == rr) return ll;
      long long mid = ((ll + rr) >> 1);
      if (l[x << 1].sum >= qq) return query(x << 1, ll, mid, qq);
      return query(x << 1 | 1, mid + 1, rr, qq - l[x << 1].sum);
    }
    long long query2(long long x, long long ll, long long rr, long long qq) {
      if (l[x].sum < qq) return -1;
      if (ll == rr) return ll;
      long long mid = ((ll + rr) >> 1);
      if (l[x << 1 | 1].sum >= qq) return query2(x << 1 | 1, mid + 1, rr, qq);
      return query2(x << 1, ll, mid, qq - l[x << 1 | 1].sum);
    }
    long long query1(long long x, long long ll, long long rr, long long qq,
                     long long ww) {
      if (qq > ww) return 0;
      if (ll >= qq && rr <= ww) return l[x].sum;
      long long mid = ((ll + rr) >> 1);
      if (mid >= ww)
        return query1(x << 1, ll, mid, qq, ww);
      else if (mid < qq)
        return query1(x << 1 | 1, mid + 1, rr, qq, ww);
      return query1(x << 1, ll, mid, qq, ww) +
             query1(x << 1 | 1, mid + 1, rr, qq, ww);
    }
    void build(long long x, long long ll, long long rr) {
      if (ll == rr) {
        l[x].sum = v1[ll];
        return;
      }
      long long mid = ((ll + rr) >> 1);
      build(x << 1, ll, mid);
      build(x << 1 | 1, mid + 1, rr);
      pushup(x);
    }
  } seg[8], s;
  void work1(long long qq, long long ww, long long ee, long long ff) {
    if (fl) return;
    long long t1 = seg[qq].query(1, 1, cn, ff);
    if (t1 == -1) return;
    long long tt = seg[ww].query1(1, 1, cn, 1, t1);
    long long t2 = seg[ww].query(1, 1, cn, ff + tt);
    if (t2 == -1) return;
    tt = seg[ee].query1(1, 1, cn, 1, t2);
    long long t3 = seg[ee].query(1, 1, cn, ff + tt);
    if (t3 == -1) return;
    fl = 1;
  }
  void work2(long long qq, long long ww, long long ee, long long ff) {
    if (fl) return;
    long long t1 = seg[qq + 3].query(1, 1, cn, ff);
    if (t1 == -1) return;
    long long tt = seg[ww + 3].query1(1, 1, cn, 1, t1);
    long long t2 = seg[ww + 3].query(1, 1, cn, ff + tt);
    if (t2 == -1) return;
    tt = seg[ee + 3].query1(1, 1, cn, 1, t2);
    long long t3 = seg[ee + 3].query(1, 1, cn, ff + tt);
    if (t3 == -1) return;
    fl = 1;
  }
  void work3(long long qq, long long ww, long long ee, long long ff) {
    if (fl) return;
    long long t1 = seg[qq].query(1, 1, cn, ff);
    if (t1 == -1) return;
    memset(v1, 0, sizeof(v1));
    for (int i = 0; i < qu[ww].size(); i++)
      if (qu[ww][i].q > t1) v1[qu[ww][i].w]++;
    s.build(1, 1, cn);
    memset(v1, 0, sizeof(v1));
    long long t2 = s.query(1, 1, cn, ff);
    if (t2 == -1) return;
    for (int i = 0; i < qu[ee].size(); i++)
      if (qu[ee][i].q > t1) v1[qu[ee][i].w]++;
    s.build(1, 1, cn);
    long long tt = s.query1(1, 1, cn, 1, t2);
    long long t3 = s.query(1, 1, cn, ff + tt);
    if (t3 == -1) return;
    fl = 1;
  }
  void work4(long long qq, long long ww, long long ee, long long ff) {
    if (fl) return;
    long long t1 = seg[qq + 3].query(1, 1, cn, ff);
    if (t1 == -1) return;
    memset(v1, 0, sizeof(v1));
    for (int i = 0; i < qu[ww].size(); i++)
      if (qu[ww][i].w > t1) v1[qu[ww][i].q]++;
    s.build(1, 1, cn);
    memset(v1, 0, sizeof(v1));
    long long t2 = s.query(1, 1, cn, ff);
    if (t2 == -1) return;
    for (int i = 0; i < qu[ee].size(); i++)
      if (qu[ee][i].w > t1) v1[qu[ee][i].q]++;
    s.build(1, 1, cn);
    long long tt = s.query1(1, 1, cn, 1, t2);
    long long t3 = s.query(1, 1, cn, ff + tt);
    if (t3 == -1) return;
    fl = 1;
  }
  void work5(long long qq, long long ww, long long ee, long long ff) {
    if (fl) return;
    long long t1 = seg[qq].query2(1, 1, cn, ff);
    if (t1 == -1) return;
    memset(v1, 0, sizeof(v1));
    for (int i = 0; i < qu[ww].size(); i++)
      if (qu[ww][i].q < t1) v1[qu[ww][i].w]++;
    s.build(1, 1, cn);
    memset(v1, 0, sizeof(v1));
    long long t2 = s.query(1, 1, cn, ff);
    if (t2 == -1) return;
    for (int i = 0; i < qu[ee].size(); i++)
      if (qu[ee][i].q < t1) v1[qu[ee][i].w]++;
    s.build(1, 1, cn);
    long long tt = s.query1(1, 1, cn, 1, t2);
    long long t3 = s.query(1, 1, cn, ff + tt);
    if (t3 == -1) return;
    fl = 1;
  }
  void work6(long long qq, long long ww, long long ee, long long ff) {
    if (fl) return;
    long long t1 = seg[qq + 3].query2(1, 1, cn, ff);
    if (t1 == -1) return;
    memset(v1, 0, sizeof(v1));
    for (int i = 0; i < qu[ww].size(); i++)
      if (qu[ww][i].w < t1) v1[qu[ww][i].q]++;
    s.build(1, 1, cn);
    memset(v1, 0, sizeof(v1));
    long long t2 = s.query(1, 1, cn, ff);
    if (t2 == -1) return;
    for (int i = 0; i < qu[ee].size(); i++)
      if (qu[ee][i].w < t1) v1[qu[ee][i].q]++;
    s.build(1, 1, cn);
    long long tt = s.query1(1, 1, cn, 1, t2);
    long long t3 = s.query(1, 1, cn, ff + tt);
    if (t3 == -1) return;
    fl = 1;
  }
  bool ch(long long qq) {
    fl = 0;
    work1(1, 2, 3, qq);
    work1(1, 3, 2, qq);
    work1(2, 1, 3, qq);
    work1(2, 3, 1, qq);
    work1(3, 1, 2, qq);
    work1(3, 2, 1, qq);
    work2(1, 2, 3, qq);
    work2(1, 3, 2, qq);
    work2(2, 1, 3, qq);
    work2(2, 3, 1, qq);
    work2(3, 1, 2, qq);
    work2(3, 2, 1, qq);
    work3(1, 2, 3, qq);
    work3(1, 3, 2, qq);
    work3(2, 1, 3, qq);
    work3(2, 3, 1, qq);
    work3(3, 1, 2, qq);
    work3(3, 2, 1, qq);
    work4(1, 2, 3, qq);
    work4(1, 3, 2, qq);
    work4(2, 1, 3, qq);
    work4(2, 3, 1, qq);
    work4(3, 1, 2, qq);
    work4(3, 2, 1, qq);
    work5(1, 2, 3, qq);
    work5(1, 3, 2, qq);
    work5(2, 1, 3, qq);
    work5(2, 3, 1, qq);
    work5(3, 1, 2, qq);
    work5(3, 2, 1, qq);
    work6(1, 2, 3, qq);
    work6(1, 3, 2, qq);
    work6(2, 1, 3, qq);
    work6(2, 3, 1, qq);
    work6(3, 1, 2, qq);
    work6(3, 2, 1, qq);
    return fl;
  }
  int main() {
    scanf("%lld", &a);
    for (int i = 1; i <= a; i++) {
      scanf("%lld%lld%lld", &q, &w, &e);
      qu[e].push_back(p{q, w, i});
      u[++cn] = q;
      u[++cn] = w;
    }
    sort(u + 1, u + cn + 1);
    cn = unique(u + 1, u + cn + 1) - u - 1;
    for (int i = 1; i <= 3; i++)
      for (int j = 0; j < qu[i].size(); j++)
        qu[i][j].q = lower_bound(u + 1, u + cn + 1, qu[i][j].q) - u,
        qu[i][j].w = lower_bound(u + 1, u + cn + 1, qu[i][j].w) - u;
    for (int i = 1; i <= 3; i++) {
      memset(v1, 0, sizeof(v1));
      memset(v2, 0, sizeof(v2));
      for (int j = 0; j < qu[i].size(); j++) v1[qu[i][j].q]++, v2[qu[i][j].w]++;
      for (int j = 1; j <= cn; j++) {
        seg[i].change(1, 1, cn, j, v1[j]);
        seg[i + 3].change(1, 1, cn, j, v2[j]);
      }
    }
    long long ll = 1, rr = a / 3;
    while (ll <= rr) {
      long long mid = ((ll + rr) >> 1);
      if (ch(mid))
        ll = mid + 1, ann = mid;
      else
        rr = mid - 1;
    }
    printf("%lld", ann * 3);
    return 0;
  }
  