#include <bits/stdc++.h>
using namespace std;
template <class T>
bool uin(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool uax(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
vector<pair<int, int> > byx[4][3], byy[4][3];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.precision(10);
  cout << fixed;
  vector<vector<pair<int, int> > > a(3);
  int n;
  cin >> n;
  for (int i = 0; i < (int)(n); ++i) {
    pair<int, int> p;
    int c;
    cin >> p.first >> p.second >> c;
    --c;
    a[c].push_back(p);
  }
  for (int z = 0; z < (int)(4); ++z) {
    for (int c = 0; c < (int)(3); ++c) {
      byx[z][c] = a[c];
      sort((byx[z][c]).begin(), (byx[z][c]).end());
      byy[z][c] = a[c];
      sort((byy[z][c]).begin(), (byy[z][c]).end(),
           [&](pair<int, int> a, pair<int, int> b) {
             return a.second < b.second;
           });
      for (auto &p : a[c]) {
        swap(p.first, p.second);
        p.second = -p.second;
      }
    }
  }
  int L = 0, R = n / 3 + 1;
  while (R - L > 1) {
    int M = (L + R) / 2;
    bool ok = false;
    for (int z = 0; z < (int)(4); ++z)
      for (int c = 0; c < (int)(3); ++c) {
        int lx = byx[z][c][M - 1].first + 1;
        vector<vector<pair<int, int> > > b(2);
        for (int d = 0; d < (int)(2); ++d) {
          int cc = (c + 1 + d) % 3;
          for (auto p : byy[z][cc])
            if (p.first >= lx) b[d].push_back(p);
        }
        if (b[0].size() < M || b[1].size() < M) continue;
        for (int z = 0; z < (int)(2); ++z)
          ok |= b[z][M - 1].second < b[z ^ 1].end()[-M].second;
        for (int d = 0; d < (int)(2); ++d) {
          int lxx = lx;
          bool good = true;
          for (int e = 0; e < (int)(2); ++e) {
            int cc = (c + 1 + (d ^ e)) % 3;
            auto it1 = lower_bound((byx[z][cc]).begin(), (byx[z][cc]).end(),
                                   make_pair(lxx, -1));
            if (byx[z][cc].end() - it1 < M) {
              good = false;
              break;
            }
            lxx = (it1 + M - 1)->first + 1;
          }
          if (good) ok = true;
        }
      }
    (ok ? L : R) = M;
  }
  cout << 3 * L << '\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;
template <class T>
bool uin(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool uax(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
vector<pair<int, int> > byx[4][3], byy[4][3];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.precision(10);
  cout << fixed;
  vector<vector<pair<int, int> > > a(3);
  int n;
  cin >> n;
  for (int i = 0; i < (int)(n); ++i) {
    pair<int, int> p;
    int c;
    cin >> p.first >> p.second >> c;
    --c;
    a[c].push_back(p);
  }
  for (int z = 0; z < (int)(4); ++z) {
    for (int c = 0; c < (int)(3); ++c) {
      byx[z][c] = a[c];
      sort((byx[z][c]).begin(), (byx[z][c]).end());
      byy[z][c] = a[c];
      sort((byy[z][c]).begin(), (byy[z][c]).end(),
           [&](pair<int, int> a, pair<int, int> b) {
             return a.second < b.second;
           });
      for (auto &p : a[c]) {
        swap(p.first, p.second);
        p.second = -p.second;
      }
    }
  }
  int L = 0, R = n / 3 + 1;
  while (R - L > 1) {
    int M = (L + R) / 2;
    bool ok = false;
    for (int z = 0; z < (int)(4); ++z)
      for (int c = 0; c < (int)(3); ++c) {
        int lx = byx[z][c][M - 1].first + 1;
        vector<vector<pair<int, int> > > b(2);
        for (int d = 0; d < (int)(2); ++d) {
          int cc = (c + 1 + d) % 3;
          for (auto p : byy[z][cc])
            if (p.first >= lx) b[d].push_back(p);
        }
        if (b[0].size() < M || b[1].size() < M) continue;
        for (int z = 0; z < (int)(2); ++z)
          ok |= b[z][M - 1].second < b[z ^ 1].end()[-M].second;
        for (int d = 0; d < (int)(2); ++d) {
          int lxx = lx;
          bool good = true;
          for (int e = 0; e < (int)(2); ++e) {
            int cc = (c + 1 + (d ^ e)) % 3;
            auto it1 = lower_bound((byx[z][cc]).begin(), (byx[z][cc]).end(),
                                   make_pair(lxx, -1));
            if (byx[z][cc].end() - it1 < M) {
              good = false;
              break;
            }
            lxx = (it1 + M - 1)->first + 1;
          }
          if (good) ok = true;
        }
      }
    (ok ? L : R) = M;
  }
  cout << 3 * L << '\n';
  return 0;
}

```