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 = 300005, MAXT = MAXN * 20;
  int n, xx[MAXN], yy[MAXN], cnt[MAXT][3], ls[MAXT], rs[MAXT];
  int rt[MAXN], per[3], chs[MAXN], nx, ny, tot, ans;
  struct Po {
    int x, y, c, id;
  } po[MAXN];
  bool cmpx(const Po &x, const Po &y) { return x.x < y.x; }
  int update(int pre, int x, int c, int l, int r) {
    int k = ++tot;
    ls[k] = ls[pre], rs[k] = rs[pre];
    memcpy(cnt[k], cnt[pre], sizeof(cnt[k]));
    ++cnt[k][c];
    if (l == r) return k;
    int mid = (l + r) >> 1;
    if (x <= mid)
      ls[k] = update(ls[pre], x, c, l, mid);
    else
      rs[k] = update(rs[pre], x, c, mid + 1, r);
    return k;
  }
  int query1(int k, int a, int b, int c, int l, int r) {
    if (!k || l > b || r < a) return 0;
    if (l >= a && r <= b) return cnt[k][c];
    int mid = (l + r) >> 1;
    return query1(ls[k], a, b, c, l, mid) + query1(rs[k], a, b, c, mid + 1, r);
  }
  int query(int x1, int y1, int x2, int y2, int c) {
    x1 = max(x1, 1);
    x2 = min(x2, nx);
    y1 = max(y1, 1);
    y2 = min(y2, ny);
    if (x1 > x2 || y1 > y2) return 0;
    return query1(rt[x2], y1, y2, c, 1, ny) -
           query1(rt[x1 - 1], y1, y2, c, 1, ny);
  }
  void setans(int x1, int y1, int x2, int y2, int c) {
    int cnt = 0;
    for (int i = 1; i <= n && cnt < ans; i++) {
      if (po[i].x >= x1 && po[i].y >= y1 && po[i].x <= x2 && po[i].y <= y2 &&
          po[i].c == c)
        chs[i] = 1, ++cnt;
    }
  }
  int cmax(int x1, int y1, int x2, int y2, int dep, bool out) {
    if (x1 > x2 || y1 > y2) return 0;
    if (dep == 2) {
      if (out) setans(x1, y1, x2, y2, per[2]);
      return query(x1, y1, x2, y2, per[2]);
    }
    int l = x1 - 1, r = x2 + 1;
    while (l + 1 < r) {
      int mid = (l + r) >> 1;
      if (query(x1, y1, mid, y2, per[dep]) <=
          cmax(mid + 1, y1, x2, y2, dep + 1, 0))
        l = mid;
      else
        r = mid;
    }
    int a = query(x1, y1, l, y2, per[dep]);
    if (out && a >= ans) {
      setans(x1, y1, l, y2, per[dep]);
      cmax(l + 1, y1, x2, y2, dep + 1, 1);
      return a;
    }
    a = max(a, cmax(l + 2, y1, x2, y2, dep + 1, 0));
    if (out && a >= ans) {
      setans(x1, y1, l + 1, y2, per[dep]);
      cmax(l + 2, y1, x2, y2, dep + 1, 1);
      return a;
    }
    l = x1 - 1, r = x2 + 1;
    while (l + 1 < r) {
      int mid = (l + r) >> 1;
      if (cmax(x1, y1, mid, y2, dep + 1, 0) <=
          query(mid + 1, y1, x2, y2, per[dep]))
        l = mid;
      else
        r = mid;
    }
    a = max(a, cmax(x1, y1, l, y2, dep + 1, 0));
    if (out && a >= ans) {
      cmax(x1, y1, l, y2, dep + 1, 1);
      setans(l + 1, y1, x2, y2, per[dep]);
      return a;
    }
    a = max(a, query(l + 2, y1, x2, y2, per[dep]));
    if (out && a >= ans) {
      cmax(x1, y1, l + 1, y2, dep + 1, 1);
      setans(l + 2, y1, x2, y2, per[dep]);
      return a;
    }
    l = y1 - 1, r = y2 + 1;
    while (l + 1 < r) {
      int mid = (l + r) >> 1;
      if (query(x1, y1, x2, mid, per[dep]) <=
          cmax(x1, mid + 1, x2, y2, dep + 1, 0))
        l = mid;
      else
        r = mid;
    }
    a = max(a, query(x1, y1, x2, l, per[dep]));
    if (out && a >= ans) {
      setans(x1, y1, x2, l, per[dep]);
      cmax(x1, l + 1, x2, y2, dep + 1, 1);
      return a;
    }
    a = max(a, cmax(x1, l + 2, x2, y2, dep + 1, 0));
    if (out && a >= ans) {
      setans(x1, y1, x2, l + 1, per[dep]);
      cmax(x1, l + 2, x2, y2, dep + 1, 1);
      return a;
    }
    l = y1 - 1, r = y2 + 1;
    while (l + 1 < r) {
      int mid = (l + r) >> 1;
      if (cmax(x1, y1, x2, mid, dep + 1, 0) <=
          query(x1, mid + 1, x2, y2, per[dep]))
        l = mid;
      else
        r = mid;
    }
    a = max(a, cmax(x1, y1, x2, l, dep + 1, 0));
    if (out && a >= ans) {
      cmax(x1, y1, x2, l, dep + 1, 1);
      setans(x1, l + 1, x2, y2, per[dep]);
      return a;
    }
    a = max(a, query(x1, l + 2, x2, y2, per[dep]));
    if (out && a >= ans) {
      cmax(x1, y1, x2, l + 1, dep + 1, 1);
      setans(x1, l + 2, x2, y2, per[dep]);
      return a;
    }
    return a;
  }
  int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
      int x, y, c;
      scanf("%d%d%d", &x, &y, &c);
      po[i] = (Po){x, y, c - 1, i};
      xx[i] = x, yy[i] = y;
    }
    sort(xx + 1, xx + 1 + n);
    sort(yy + 1, yy + 1 + n);
    nx = unique(xx + 1, xx + 1 + n) - xx - 1;
    ny = unique(yy + 1, yy + 1 + n) - yy - 1;
    sort(po + 1, po + 1 + n, cmpx);
    for (int i = 1; i <= n; i++) {
      po[i].x = lower_bound(xx + 1, xx + nx + 1, po[i].x) - xx;
      po[i].y = lower_bound(yy + 1, yy + ny + 1, po[i].y) - yy;
      rt[po[i].x] = update(rt[po[i - 1].x], po[i].y, po[i].c, 1, ny);
    }
    per[0] = 0, per[1] = 1, per[2] = 2;
    do {
      ans = max(ans, cmax(1, 1, nx, ny, 0, 0));
    } while (next_permutation(per, per + 3));
    per[0] = 0, per[1] = 1, per[2] = 2;
    do {
      if (cmax(1, 1, nx, ny, 0, 1) >= ans) break;
    } while (next_permutation(per, per + 3));
    printf("%d\n", ans * 3);
    return 0;
  }
  