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;
  namespace jyy {
  const int mod = 998244353, N = 400010;
  const double eps = 1e-8;
  inline int read() {
    static int x = 0, f = 1;
    x = 0, f = 1;
    static char ch = getchar();
    while (ch < '0' || ch > '9') {
      (ch == '-') ? f *= -1 : f;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') x = x * 10 + int(ch - 48), ch = getchar();
    return x * f;
  }
  inline void fw(int x) {
    if (x < 0) putchar('-'), x = -x;
    x >= 10 ? fw(x / 10), 1 : 0;
    putchar(char(x % 10 + 48));
  }
  inline int qmo(int x) { return x + ((x >> 31) & mod); }
  inline int ksm(int x, int y) {
    int tmp = 1;
    for (; y; y >>= 1, x = 1ll * x * x % mod)
      if (y & 1) tmp = 1ll * tmp * x % mod;
    return tmp;
  }
  inline int inv(int x) { return ksm(x, mod - 2); }
  int tot, head[N], fa[N];
  struct edge {
    int to, next;
  } dat[N * 2];
  void add_edge(int x, int y) { dat[++tot] = (edge){y, head[x]}, head[x] = tot; }
  inline int find(int x) { return fa[x] ? fa[x] = find(fa[x]) : x; }
  inline int merge(int x, int y) {
    return find(x) != find(y) ? fa[find(x)] = find(y), 0 : 1;
  }
  inline int low(int x) { return x & -x; }
  int ifac[N], fac[N];
  inline void init(int x) {
    fac[0] = ifac[0] = 1;
    for (int i = 1; i <= x; i++) fac[i] = 1ll * fac[i - 1] * i % mod;
    ifac[x] = inv(fac[x]);
    for (int i = x - 1; i >= 1; i--) ifac[i] = 1ll * ifac[i + 1] * (i + 1) % mod;
  }
  inline int C(int x, int y) {
    return (x < y || y < 0 || x < 0)
               ? 0
               : 1ll * fac[x] * ifac[y] % mod * ifac[x - y] % mod;
  }
  }  // namespace jyy
  using namespace jyy;
  int a[N], n, b[N], c[N], p[N], cnt, val[N];
  vector<int> v1[4][N], v2[4][N];
  int do11(int mid) {
    int num = 0, now = 0;
    for (int i = 1; i <= cnt; i++) {
      now = i + 1;
      num += v1[p[1]][i].size();
      if (num >= mid) break;
    }
    if (now > cnt) return 0;
    int hh = now;
    num = 0;
    for (int i = now; i <= cnt; i++) {
      now = i + 1;
      num += v1[p[2]][i].size();
      if (num >= mid) break;
    }
    if (now > cnt) return 0;
    num = 0;
    for (int i = now; i <= cnt; i++) num += v1[p[3]][i].size();
    if (num >= mid) return 1;
    num = 0;
    now = 0;
    for (int i = 1; i <= cnt; i++) {
      now = i + 1;
      num += v2[p[2]][i].end() -
             lower_bound(v2[p[2]][i].begin(), v2[p[2]][i].end(), hh);
      if (num >= mid) break;
    }
    if (now > cnt) return 0;
    num = 0;
    for (int i = now; i <= cnt; i++) {
      num += v2[p[3]][i].end() -
             lower_bound(v2[p[3]][i].begin(), v2[p[3]][i].end(), hh);
    }
    if (num >= mid) return 1;
    return 0;
  }
  int do21(int mid) {
    int num = 0, now = cnt + 1;
    for (int i = cnt; i >= 1; i--) {
      now = i - 1;
      num += v1[p[1]][i].size();
      if (num >= mid) break;
    }
    if (!now) return 0;
    int hh = now;
    num = 0;
    now = 0;
    for (int i = 1; i <= cnt; i++) {
      now = i + 1;
      num += lower_bound(v2[p[2]][i].begin(), v2[p[2]][i].end(), hh + 1) -
             v2[p[2]][i].begin();
      if (num >= mid) break;
    }
    if (now > cnt) return 0;
    num = 0;
    for (int i = now; i <= cnt; i++) {
      num += lower_bound(v2[p[3]][i].begin(), v2[p[3]][i].end(), hh + 1) -
             v2[p[3]][i].begin();
    }
    if (num >= mid) return 1;
    return 0;
  }
  int do22(int mid) {
    int num = 0, now = cnt + 1;
    for (int i = cnt; i >= 1; i--) {
      now = i - 1;
      num += v2[p[1]][i].size();
      if (num >= mid) break;
    }
    if (!now) return 0;
    int hh = now;
    num = 0;
    now = 0;
    for (int i = 1; i <= cnt; i++) {
      now = i + 1;
      num += lower_bound(v1[p[2]][i].begin(), v1[p[2]][i].end(), hh + 1) -
             v1[p[2]][i].begin();
      if (num >= mid) break;
    }
    if (now > cnt) return 0;
    num = 0;
    for (int i = now; i <= cnt; i++) {
      num += lower_bound(v1[p[3]][i].begin(), v1[p[3]][i].end(), hh + 1) -
             v1[p[3]][i].begin();
    }
    if (num >= mid) return 1;
    return 0;
  }
  int do12(int mid) {
    int num = 0, now = 0;
    for (int i = 1; i <= cnt; i++) {
      now = i + 1;
      num += v2[p[1]][i].size();
      if (num >= mid) break;
    }
    if (now > cnt) return 0;
    int hh = now;
    num = 0;
    for (int i = now; i <= cnt; i++) {
      now = i + 1;
      num += v2[p[2]][i].size();
      if (num >= mid) break;
    }
    if (now > cnt) return 0;
    num = 0;
    for (int i = now; i <= cnt; i++) num += v2[p[3]][i].size();
    if (num >= mid) return 1;
    num = 0;
    now = 0;
    for (int i = 1; i <= cnt; i++) {
      now = i + 1;
      num += v1[p[2]][i].end() -
             lower_bound(v1[p[2]][i].begin(), v1[p[2]][i].end(), hh);
      if (num >= mid) break;
    }
    if (now > cnt) return 0;
    num = 0;
    for (int i = now; i <= cnt; i++) {
      num += v1[p[3]][i].end() -
             lower_bound(v1[p[3]][i].begin(), v1[p[3]][i].end(), hh);
    }
    if (num >= mid) return 1;
    return 0;
  }
  bool check(int mid) {
    for (int i = 1; i <= 3; i++) p[i] = i;
    do {
      if (do11(mid)) return 1;
      if (do12(mid)) return 1;
      if (do21(mid)) return 1;
      if (do22(mid)) return 1;
    } while (next_permutation(p + 1, p + 1 + 3));
    return 0;
  }
  void solve() {
    n = read();
    for (int i = 1; i <= n; i++)
      a[i] = read(), b[i] = read(), c[i] = read(), val[++cnt] = a[i],
      val[++cnt] = b[i];
    sort(val + 1, val + 1 + cnt), cnt = unique(val + 1, val + 1 + cnt) - val - 1;
    for (int i = 1; i <= n; i++) {
      a[i] = lower_bound(val + 1, val + 1 + cnt, a[i]) - val;
      b[i] = lower_bound(val + 1, val + 1 + cnt, b[i]) - val;
      v1[c[i]][a[i]].push_back(b[i]);
      v2[c[i]][b[i]].push_back(a[i]);
    }
    for (int k = 1; k <= 3; k++)
      for (int i = 1; i <= cnt; i++) {
        sort(v1[k][i].begin(), v1[k][i].end());
        sort(v2[k][i].begin(), v2[k][i].end());
      }
    int l = 0, r = n / 3, res = 0;
    while (l <= r) {
      int mid = (l + r) / 2;
      if (check(mid))
        l = mid + 1, res = mid;
      else
        r = mid - 1;
    }
    cout << res * 3 << endl;
  }
  int main() {
    int T = 1;
    while (T--) solve();
  }
  