#include <bits/stdc++.h>
using namespace std;
const int CL = 3;
const int INF = 1e9 + 123;
vector<pair<int, int> > V[4][CL];
bool checkrow(vector<int> perm, int need, int ind) {
  int rx = -INF;
  bool flag = false;
  for (int col : perm) {
    ++rx;
    int cnt = 0;
    for (auto [x, y] : V[ind][col]) {
      if (x >= rx) ++cnt;
      rx = max(x, rx);
      if (cnt == need) break;
    }
    if (cnt != need) return false;
  }
  return true;
}
bool check1(vector<int> perm, int need) {
  bool ans = false;
  for (int u = 0; u < 2; ++u) {
    ans = max(ans, checkrow(perm, need, u));
  }
  return ans;
}
bool mc(pair<int, int> a, pair<int, int> b) { return a.second < b.second; }
bool checkT(vector<int> perm, int need, int ind) {
  int rx = -INF;
  int cnt = 0;
  for (auto [x, y] : V[ind][perm[0]]) {
    rx = max(rx, x);
    ++cnt;
    if (cnt == need) break;
  }
  if (cnt != need) return false;
  ++rx;
  int ry = -INF;
  for (int j = 1; j < CL; ++j) {
    int col = perm[j];
    ++ry;
    int cnt = 0;
    for (auto [y, x] : V[(ind & 1) ^ 1][col]) {
      int sign = (ind & 2) ? -1 : 1;
      if (x * sign < rx || y < ry) continue;
      ry = max(ry, y);
      ++cnt;
      if (cnt == need) break;
    }
    if (cnt != need) return false;
  }
  return true;
}
bool check2(vector<int> perm, int need) {
  bool ans = false;
  for (int u = 0; u < 2; ++u) {
    for (int u1 = 0; u1 < 2; ++u1) {
      ans = max(ans, checkT(perm, need, u + u1 * 2));
    }
  }
  return ans;
}
bool F(int need) {
  vector<int> perm(CL);
  for (int i = 0; i < CL; ++i) perm[i] = i;
  do {
    if (check1(perm, need)) return true;
    if (check2(perm, need)) return true;
  } while (next_permutation(perm.begin(), perm.end()));
  return false;
}
void solve() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    int x, y, c;
    cin >> x >> y >> c;
    c--;
    V[0][c].push_back({x, y});
  }
  for (int c = 0; c < CL; ++c) {
    V[1][c] = V[0][c];
    V[2][c] = V[0][c];
    V[3][c] = V[0][c];
    int sz = V[0][c].size();
    for (int j = 0; j < sz; ++j)
      swap(V[1][c][j].first, V[1][c][j].second),
          swap(V[3][c][j].first, V[3][c][j].second);
    for (int j = 0; j < sz; ++j) V[2][c][j].first *= -1, V[3][c][j].first *= -1;
    for (int cl = 0; cl < 4; ++cl) sort(V[cl][c].begin(), V[cl][c].end());
  }
  int l = 0, r = n / 3;
  while (l != r) {
    int mid = (l + r + 1) >> 1;
    if (F(mid))
      l = mid;
    else
      r = mid - 1;
  }
  cout << l * 3 << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  while (t--) solve();
  return 0;
}

### Prompt

Develop a solution in CPP 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;
const int CL = 3;
const int INF = 1e9 + 123;
vector<pair<int, int> > V[4][CL];
bool checkrow(vector<int> perm, int need, int ind) {
  int rx = -INF;
  bool flag = false;
  for (int col : perm) {
    ++rx;
    int cnt = 0;
    for (auto [x, y] : V[ind][col]) {
      if (x >= rx) ++cnt;
      rx = max(x, rx);
      if (cnt == need) break;
    }
    if (cnt != need) return false;
  }
  return true;
}
bool check1(vector<int> perm, int need) {
  bool ans = false;
  for (int u = 0; u < 2; ++u) {
    ans = max(ans, checkrow(perm, need, u));
  }
  return ans;
}
bool mc(pair<int, int> a, pair<int, int> b) { return a.second < b.second; }
bool checkT(vector<int> perm, int need, int ind) {
  int rx = -INF;
  int cnt = 0;
  for (auto [x, y] : V[ind][perm[0]]) {
    rx = max(rx, x);
    ++cnt;
    if (cnt == need) break;
  }
  if (cnt != need) return false;
  ++rx;
  int ry = -INF;
  for (int j = 1; j < CL; ++j) {
    int col = perm[j];
    ++ry;
    int cnt = 0;
    for (auto [y, x] : V[(ind & 1) ^ 1][col]) {
      int sign = (ind & 2) ? -1 : 1;
      if (x * sign < rx || y < ry) continue;
      ry = max(ry, y);
      ++cnt;
      if (cnt == need) break;
    }
    if (cnt != need) return false;
  }
  return true;
}
bool check2(vector<int> perm, int need) {
  bool ans = false;
  for (int u = 0; u < 2; ++u) {
    for (int u1 = 0; u1 < 2; ++u1) {
      ans = max(ans, checkT(perm, need, u + u1 * 2));
    }
  }
  return ans;
}
bool F(int need) {
  vector<int> perm(CL);
  for (int i = 0; i < CL; ++i) perm[i] = i;
  do {
    if (check1(perm, need)) return true;
    if (check2(perm, need)) return true;
  } while (next_permutation(perm.begin(), perm.end()));
  return false;
}
void solve() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    int x, y, c;
    cin >> x >> y >> c;
    c--;
    V[0][c].push_back({x, y});
  }
  for (int c = 0; c < CL; ++c) {
    V[1][c] = V[0][c];
    V[2][c] = V[0][c];
    V[3][c] = V[0][c];
    int sz = V[0][c].size();
    for (int j = 0; j < sz; ++j)
      swap(V[1][c][j].first, V[1][c][j].second),
          swap(V[3][c][j].first, V[3][c][j].second);
    for (int j = 0; j < sz; ++j) V[2][c][j].first *= -1, V[3][c][j].first *= -1;
    for (int cl = 0; cl < 4; ++cl) sort(V[cl][c].begin(), V[cl][c].end());
  }
  int l = 0, r = n / 3;
  while (l != r) {
    int mid = (l + r + 1) >> 1;
    if (F(mid))
      l = mid;
    else
      r = mid - 1;
  }
  cout << l * 3 << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  while (t--) solve();
  return 0;
}

```