#include <bits/stdc++.h>
using namespace std;
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
void ran() { srand(chrono::steady_clock::now().time_since_epoch().count()); }
long long get_rand() {
  long long a = rand();
  long long b = rand();
  return a * (RAND_MAX + 1ll) + b;
}
void usaco() {
  freopen("problem.in", "r", stdin);
  freopen("problem.out", "w", stdout);
}
template <typename T>
using min_pq = priority_queue<T, vector<T>, greater<T>>;
int n;
const int MXN = 100005;
int xco[MXN], yco[MXN], col[MXN];
int ans() {
  int r = 0;
  vector<int> cols = {1, 2, 3};
  map<int, vector<int>> lr;
  map<int, vector<pair<int, int>>> td;
  for (int i = 1; i <= n; i++)
    lr[xco[i]].push_back(col[i]), td[yco[i]].push_back({xco[i], col[i]});
  vector<vector<int>> tr;
  vector<vector<pair<int, int>>> le;
  for (auto x : lr) tr.push_back(x.second);
  for (auto x : td) le.push_back(x.second);
  do {
    int L = cols[0], M = cols[1], R = cols[2];
    int midleft = n / 3;
    int leftf = 0, rightf = 0;
    int lpt = 0, rpt = tr.size() - 1;
    for (int k = 1; k <= n / 3; k++) {
      while (lpt < tr.size() && leftf < k) {
        for (auto m : tr[lpt]) {
          if (m == L) {
            leftf++;
          } else if (m == M) {
            midleft--;
          }
        }
        lpt++;
      }
      while (rpt >= 0 && rightf < k) {
        for (auto m : tr[rpt]) {
          if (m == R) {
            rightf++;
          } else if (m == M) {
            midleft--;
          }
        }
        rpt--;
      }
      if (leftf >= k && rightf >= k && midleft >= k) r = max(r, k);
    }
    midleft = n / 3;
    lpt = 0;
    int inbucket = INT_MIN / 4;
    min_pq<int> rightitems;
    priority_queue<int> leftitems;
    for (int k = n / 3; k >= 1; k--) {
      while (lpt < le.size()) {
        if (inbucket == INT_MIN / 4) {
          inbucket = 0;
          for (auto j : le[lpt]) {
            if (j.second == M) {
              inbucket++;
            }
          }
        }
        if (midleft - inbucket < k) {
          break;
        }
        for (auto j : le[lpt]) {
          if (j.second == L) leftitems.push(j.first);
          if (j.second == R) rightitems.push(j.first);
        }
        midleft -= inbucket;
        lpt++;
        inbucket = INT_MIN / 4;
      }
      while (leftitems.size() > k) leftitems.pop();
      while (rightitems.size() > k) rightitems.pop();
      if (leftitems.size() == k && rightitems.size() == k &&
          leftitems.top() < rightitems.top())
        r = max(r, k);
    }
  } while (next_permutation(cols.begin(), cols.end()));
  return r;
}
signed main() {
  ran();
  fast();
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> xco[i] >> yco[i] >> col[i];
  }
  int r = 0;
  for (int j = 0; j < 4; j++) {
    r = max(r, ans());
    for (int i = 1; i <= n; i++) {
      swap(xco[i], yco[i]);
      xco[i] *= -1;
    }
  }
  cout << r * 3 << '\n';
}

### 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;
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
void ran() { srand(chrono::steady_clock::now().time_since_epoch().count()); }
long long get_rand() {
  long long a = rand();
  long long b = rand();
  return a * (RAND_MAX + 1ll) + b;
}
void usaco() {
  freopen("problem.in", "r", stdin);
  freopen("problem.out", "w", stdout);
}
template <typename T>
using min_pq = priority_queue<T, vector<T>, greater<T>>;
int n;
const int MXN = 100005;
int xco[MXN], yco[MXN], col[MXN];
int ans() {
  int r = 0;
  vector<int> cols = {1, 2, 3};
  map<int, vector<int>> lr;
  map<int, vector<pair<int, int>>> td;
  for (int i = 1; i <= n; i++)
    lr[xco[i]].push_back(col[i]), td[yco[i]].push_back({xco[i], col[i]});
  vector<vector<int>> tr;
  vector<vector<pair<int, int>>> le;
  for (auto x : lr) tr.push_back(x.second);
  for (auto x : td) le.push_back(x.second);
  do {
    int L = cols[0], M = cols[1], R = cols[2];
    int midleft = n / 3;
    int leftf = 0, rightf = 0;
    int lpt = 0, rpt = tr.size() - 1;
    for (int k = 1; k <= n / 3; k++) {
      while (lpt < tr.size() && leftf < k) {
        for (auto m : tr[lpt]) {
          if (m == L) {
            leftf++;
          } else if (m == M) {
            midleft--;
          }
        }
        lpt++;
      }
      while (rpt >= 0 && rightf < k) {
        for (auto m : tr[rpt]) {
          if (m == R) {
            rightf++;
          } else if (m == M) {
            midleft--;
          }
        }
        rpt--;
      }
      if (leftf >= k && rightf >= k && midleft >= k) r = max(r, k);
    }
    midleft = n / 3;
    lpt = 0;
    int inbucket = INT_MIN / 4;
    min_pq<int> rightitems;
    priority_queue<int> leftitems;
    for (int k = n / 3; k >= 1; k--) {
      while (lpt < le.size()) {
        if (inbucket == INT_MIN / 4) {
          inbucket = 0;
          for (auto j : le[lpt]) {
            if (j.second == M) {
              inbucket++;
            }
          }
        }
        if (midleft - inbucket < k) {
          break;
        }
        for (auto j : le[lpt]) {
          if (j.second == L) leftitems.push(j.first);
          if (j.second == R) rightitems.push(j.first);
        }
        midleft -= inbucket;
        lpt++;
        inbucket = INT_MIN / 4;
      }
      while (leftitems.size() > k) leftitems.pop();
      while (rightitems.size() > k) rightitems.pop();
      if (leftitems.size() == k && rightitems.size() == k &&
          leftitems.top() < rightitems.top())
        r = max(r, k);
    }
  } while (next_permutation(cols.begin(), cols.end()));
  return r;
}
signed main() {
  ran();
  fast();
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> xco[i] >> yco[i] >> col[i];
  }
  int r = 0;
  for (int j = 0; j < 4; j++) {
    r = max(r, ans());
    for (int i = 1; i <= n; i++) {
      swap(xco[i], yco[i]);
      xco[i] *= -1;
    }
  }
  cout << r * 3 << '\n';
}

```