#include <bits/stdc++.h>
using namespace std;
int n, sok = 1e9 + 1;
vector<pair<int, int> > xr[3], yr[3];
bool solve(int mid, int ba, int ja, int bf, int jf) {
  if (ba == ja || bf == jf) {
    int xa = -sok, xf = sok;
    if (ba == ja) {
      xa = xr[ba][mid - 1].first;
    }
    if (bf == jf) {
      xf = xr[jf][n / 3 - mid].first;
    }
    if (ba == ja && bf == jf) {
      int db = 0, h = 3 - ba - bf;
      for (auto ert : xr[h]) {
        if (xa < ert.first && ert.first < xf) db++;
      }
      return (db >= mid);
    } else {
      int bal = (ba == ja ? bf : ba), jobb = (ba == ja ? jf : ja);
      int bert = -sok, bdb = 0, jert = sok, jdb = 0;
      for (auto ert : yr[bal]) {
        if (xa < ert.second && ert.second < xf) {
          bdb++;
          bert = ert.first;
          if (bdb == mid) {
            break;
          }
        }
      }
      for (int i = n / 3 - 1; i >= 0; i--) {
        auto ert = yr[jobb][i];
        if (xa < ert.second && ert.second < xf) {
          jdb++;
          jert = ert.first;
          if (jdb == mid) {
            break;
          }
        }
      }
      if (bdb == mid && jdb == mid && bert < jert) {
        return true;
      }
      return false;
    }
  }
  if (ba == bf || ja == jf) {
    int ya = -sok, yf = sok;
    if (ba == bf) {
      ya = yr[ba][mid - 1].first;
    }
    if (ja == jf) {
      yf = yr[jf][n / 3 - mid].first;
    }
    if (ba == bf && ja == jf) {
      int db = 0, h = 3 - ba - ja;
      for (auto ert : yr[h]) {
        if (ya < ert.first && ert.first < yf) db++;
      }
      return (db >= mid);
    } else {
      int kis = (ba == bf ? ja : ba), nagy = (ba == bf ? jf : bf);
      int kert = -sok, kdb = 0, nert = sok, ndb = 0;
      for (auto ert : xr[kis]) {
        if (ya < ert.second && ert.second < yf) {
          kdb++;
          kert = ert.first;
          if (kdb == mid) {
            break;
          }
        }
      }
      for (int i = n / 3 - 1; i >= 0; i--) {
        auto ert = xr[nagy][i];
        if (ya < ert.second && ert.second < yf) {
          ndb++;
          nert = ert.first;
          if (ndb == mid) {
            break;
          }
        }
      }
      if (kdb == mid && ndb == mid && kert < nert) {
        return true;
      }
      return false;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    c--;
    xr[c].push_back({a, b});
    yr[c].push_back({b, a});
  }
  for (int i = 0; i < 3; i++) {
    sort(xr[i].begin(), xr[i].end());
    sort(yr[i].begin(), yr[i].end());
  }
  int lo = 0, hi = n / 3 + 1, mid;
  while (hi - lo > 1) {
    mid = (hi + lo) / 2;
    bool jo = 0;
    for (int ba = 0; ba < 3; ba++) {
      for (int ja = 0; ja < 3; ja++) {
        for (int bf = 0; bf < 3; bf++) {
          for (int jf = 0; jf < 3; jf++) {
            if (ba == jf || ja == bf) continue;
            if (solve(mid, ba, ja, bf, jf)) {
              jo = 1;
            }
          }
        }
      }
    }
    if (jo) {
      lo = mid;
    } else {
      hi = mid;
    }
  }
  cout << 3 * lo << "\n";
  return 0;
}

### Prompt

Create a solution in cpp for the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int n, sok = 1e9 + 1;
vector<pair<int, int> > xr[3], yr[3];
bool solve(int mid, int ba, int ja, int bf, int jf) {
  if (ba == ja || bf == jf) {
    int xa = -sok, xf = sok;
    if (ba == ja) {
      xa = xr[ba][mid - 1].first;
    }
    if (bf == jf) {
      xf = xr[jf][n / 3 - mid].first;
    }
    if (ba == ja && bf == jf) {
      int db = 0, h = 3 - ba - bf;
      for (auto ert : xr[h]) {
        if (xa < ert.first && ert.first < xf) db++;
      }
      return (db >= mid);
    } else {
      int bal = (ba == ja ? bf : ba), jobb = (ba == ja ? jf : ja);
      int bert = -sok, bdb = 0, jert = sok, jdb = 0;
      for (auto ert : yr[bal]) {
        if (xa < ert.second && ert.second < xf) {
          bdb++;
          bert = ert.first;
          if (bdb == mid) {
            break;
          }
        }
      }
      for (int i = n / 3 - 1; i >= 0; i--) {
        auto ert = yr[jobb][i];
        if (xa < ert.second && ert.second < xf) {
          jdb++;
          jert = ert.first;
          if (jdb == mid) {
            break;
          }
        }
      }
      if (bdb == mid && jdb == mid && bert < jert) {
        return true;
      }
      return false;
    }
  }
  if (ba == bf || ja == jf) {
    int ya = -sok, yf = sok;
    if (ba == bf) {
      ya = yr[ba][mid - 1].first;
    }
    if (ja == jf) {
      yf = yr[jf][n / 3 - mid].first;
    }
    if (ba == bf && ja == jf) {
      int db = 0, h = 3 - ba - ja;
      for (auto ert : yr[h]) {
        if (ya < ert.first && ert.first < yf) db++;
      }
      return (db >= mid);
    } else {
      int kis = (ba == bf ? ja : ba), nagy = (ba == bf ? jf : bf);
      int kert = -sok, kdb = 0, nert = sok, ndb = 0;
      for (auto ert : xr[kis]) {
        if (ya < ert.second && ert.second < yf) {
          kdb++;
          kert = ert.first;
          if (kdb == mid) {
            break;
          }
        }
      }
      for (int i = n / 3 - 1; i >= 0; i--) {
        auto ert = xr[nagy][i];
        if (ya < ert.second && ert.second < yf) {
          ndb++;
          nert = ert.first;
          if (ndb == mid) {
            break;
          }
        }
      }
      if (kdb == mid && ndb == mid && kert < nert) {
        return true;
      }
      return false;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    c--;
    xr[c].push_back({a, b});
    yr[c].push_back({b, a});
  }
  for (int i = 0; i < 3; i++) {
    sort(xr[i].begin(), xr[i].end());
    sort(yr[i].begin(), yr[i].end());
  }
  int lo = 0, hi = n / 3 + 1, mid;
  while (hi - lo > 1) {
    mid = (hi + lo) / 2;
    bool jo = 0;
    for (int ba = 0; ba < 3; ba++) {
      for (int ja = 0; ja < 3; ja++) {
        for (int bf = 0; bf < 3; bf++) {
          for (int jf = 0; jf < 3; jf++) {
            if (ba == jf || ja == bf) continue;
            if (solve(mid, ba, ja, bf, jf)) {
              jo = 1;
            }
          }
        }
      }
    }
    if (jo) {
      lo = mid;
    } else {
      hi = mid;
    }
  }
  cout << 3 * lo << "\n";
  return 0;
}

```