#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
vector<vector<pair<int, int>>> p;
vector<int> ord = {0, 1, 2};
vector<vector<vector<pair<int, int>>>> v(3), v2(3);
bool solveLine(int k) {
  int lst = -2e9;
  for (int i : ord) {
    bool found = false;
    for (int j = 0; j < p[i].size(); j++) {
      if (p[i][j].first > lst) {
        if (j + k - 1 >= p[i].size()) return false;
        lst = p[i][j + k - 1].first;
        found = true;
        break;
      }
    }
    if (!found) return false;
  }
  return true;
}
bool solveGroup(int k) {
  int to = p[ord[0]][k - 1].second;
  vector<int> a, b;
  for (int i = 0; i < p[1].size(); i++) {
    if (p[ord[1]][i].second > to) a.push_back(p[ord[1]][i].first);
    if (p[ord[2]][i].second > to) b.push_back(p[ord[2]][i].first);
  }
  int lst = -2e9;
  bool found = false;
  for (int j = 0; j < a.size(); j++) {
    if (a[j] > lst) {
      if (j + k - 1 >= a.size()) return false;
      lst = a[j + k - 1];
      found = true;
      break;
    }
  }
  if (!found) return false;
  for (int j = 0; j < b.size(); j++) {
    if (b[j] > lst) {
      if (j + k - 1 >= b.size()) return false;
      lst = b[j + k - 1];
      return true;
    }
  }
  return false;
}
bool can(int k) {
  sort((ord).begin(), (ord).end());
  do {
    int cnt = 0;
    for (int t = 0; t < 2; t++) {
      for (int i = 0; i < 2; i++) {
        for (int j = 0; j < 2; j++) {
          for (int x = 0; x < 3; x++) p[x] = v[x][cnt];
          bool res = solveLine(k);
          p[ord[0]] = v2[ord[0]][cnt];
          cnt++;
          res |= solveGroup(k);
          if (res) return true;
        }
      }
    }
  } while (next_permutation((ord).begin(), (ord).end()));
  return false;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  p.resize(3);
  for (int i = 0; i < n; i++) {
    int x, y, c;
    cin >> x >> y >> c;
    c--;
    p[c].push_back({x, y});
  }
  for (int t = 0; t < 2; t++) {
    for (int x = 0; x < 3; x++)
      for (int y = 0; y < p[x].size(); y++) swap(p[x][y].first, p[x][y].second);
    for (int i = 0; i < 2; i++) {
      for (int x = 0; x < 3; x++)
        for (int y = 0; y < p[x].size(); y++) p[x][y].first *= -1;
      for (int j = 0; j < 2; j++) {
        for (int x = 0; x < 3; x++)
          for (int y = 0; y < p[x].size(); y++) p[x][y].second *= -1;
        for (int x = 0; x < 3; x++) {
          v[x].push_back(p[x]);
          v2[x].push_back(p[x]);
          sort((v[x].back()).begin(), (v[x].back()).end());
          sort((v2[x].back()).begin(), (v2[x].back()).end(),
               [](pair<int, int> a, pair<int, int> b) {
                 return a.second < b.second;
               });
        }
      }
    }
  }
  int lo = 1, hi = n / 3, ans = 1;
  while (lo <= hi) {
    int md = (lo + hi) / 2;
    if (can(md)) {
      lo = md + 1;
      ans = md;
    } else
      hi = md - 1;
  }
  cout << ans * 3 << '\n';
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
vector<vector<pair<int, int>>> p;
vector<int> ord = {0, 1, 2};
vector<vector<vector<pair<int, int>>>> v(3), v2(3);
bool solveLine(int k) {
  int lst = -2e9;
  for (int i : ord) {
    bool found = false;
    for (int j = 0; j < p[i].size(); j++) {
      if (p[i][j].first > lst) {
        if (j + k - 1 >= p[i].size()) return false;
        lst = p[i][j + k - 1].first;
        found = true;
        break;
      }
    }
    if (!found) return false;
  }
  return true;
}
bool solveGroup(int k) {
  int to = p[ord[0]][k - 1].second;
  vector<int> a, b;
  for (int i = 0; i < p[1].size(); i++) {
    if (p[ord[1]][i].second > to) a.push_back(p[ord[1]][i].first);
    if (p[ord[2]][i].second > to) b.push_back(p[ord[2]][i].first);
  }
  int lst = -2e9;
  bool found = false;
  for (int j = 0; j < a.size(); j++) {
    if (a[j] > lst) {
      if (j + k - 1 >= a.size()) return false;
      lst = a[j + k - 1];
      found = true;
      break;
    }
  }
  if (!found) return false;
  for (int j = 0; j < b.size(); j++) {
    if (b[j] > lst) {
      if (j + k - 1 >= b.size()) return false;
      lst = b[j + k - 1];
      return true;
    }
  }
  return false;
}
bool can(int k) {
  sort((ord).begin(), (ord).end());
  do {
    int cnt = 0;
    for (int t = 0; t < 2; t++) {
      for (int i = 0; i < 2; i++) {
        for (int j = 0; j < 2; j++) {
          for (int x = 0; x < 3; x++) p[x] = v[x][cnt];
          bool res = solveLine(k);
          p[ord[0]] = v2[ord[0]][cnt];
          cnt++;
          res |= solveGroup(k);
          if (res) return true;
        }
      }
    }
  } while (next_permutation((ord).begin(), (ord).end()));
  return false;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  p.resize(3);
  for (int i = 0; i < n; i++) {
    int x, y, c;
    cin >> x >> y >> c;
    c--;
    p[c].push_back({x, y});
  }
  for (int t = 0; t < 2; t++) {
    for (int x = 0; x < 3; x++)
      for (int y = 0; y < p[x].size(); y++) swap(p[x][y].first, p[x][y].second);
    for (int i = 0; i < 2; i++) {
      for (int x = 0; x < 3; x++)
        for (int y = 0; y < p[x].size(); y++) p[x][y].first *= -1;
      for (int j = 0; j < 2; j++) {
        for (int x = 0; x < 3; x++)
          for (int y = 0; y < p[x].size(); y++) p[x][y].second *= -1;
        for (int x = 0; x < 3; x++) {
          v[x].push_back(p[x]);
          v2[x].push_back(p[x]);
          sort((v[x].back()).begin(), (v[x].back()).end());
          sort((v2[x].back()).begin(), (v2[x].back()).end(),
               [](pair<int, int> a, pair<int, int> b) {
                 return a.second < b.second;
               });
        }
      }
    }
  }
  int lo = 1, hi = n / 3, ans = 1;
  while (lo <= hi) {
    int md = (lo + hi) / 2;
    if (can(md)) {
      lo = md + 1;
      ans = md;
    } else
      hi = md - 1;
  }
  cout << ans * 3 << '\n';
}

```