#include <bits/stdc++.h>
int n;
std::vector<std::pair<int, int> > B[3], A[8][3];
inline bool chk(int x) {
  int p[3] = {0, 1, 2};
  do {
    for (int o = 0; o < 4; o++) {
      int u = A[o << 1][p[0]][x - 1].first + 1;
      int y = x, v = ((int)1e9 + 1);
      for (auto [i, j] : A[o << 1][p[1]])
        if (i >= u && !--y) {
          v = i + 1;
          break;
        }
      y = x;
      for (auto [i, j] : A[o << 1][p[2]])
        if (i >= v && !--y) return 1;
      y = x, v = ((int)1e9 + 1);
      for (auto [i, j] : A[o << 1 | 1][p[1]])
        if (i >= u && !--y) {
          v = j + 1;
          break;
        }
      y = x;
      for (auto [i, j] : A[o << 1 | 1][p[2]])
        if (i >= u && j >= v && !--y) return 1;
    }
  } while (std::next_permutation(p, p + 3));
  return 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x, y, c;
    scanf("%d%d%d", &x, &y, &c);
    B[c - 1].emplace_back(x, y);
  }
  for (int oo = 0; oo < 2; oo++) {
    for (int o = 0; o < 2; o++) {
      for (int i = 0; i < 3; i++) {
        A[(oo << 2) | (o << 1)][i] = B[i];
        std::sort(A[(oo << 2) | (o << 1)][i].begin(),
                  A[(oo << 2) | (o << 1)][i].end());
        A[(oo << 2) | (o << 1) | 1][i] = B[i];
        std::sort(A[(oo << 2) | (o << 1) | 1][i].begin(),
                  A[(oo << 2) | (o << 1) | 1][i].end(),
                  [](std::pair<int, int> u, std::pair<int, int> v) {
                    return u.second == v.second ? u.first < v.first
                                                : u.second < v.second;
                  });
      }
      for (int i = 0; i < 3; i++)
        for (int j = 0; j < n / 3; j++) B[i][j].first *= -1;
    }
    for (int i = 0; i < 3; i++)
      for (int j = 0; j < n / 3; j++) std::swap(B[i][j].first, B[i][j].second);
  }
  int L = 1, R = n / 3, res = L;
  while (L <= R) {
    int mid = (L + R) >> 1;
    if (chk(mid))
      res = mid, L = mid + 1;
    else
      R = mid - 1;
  }
  printf("%d\n", 3 * res);
}

### 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>
int n;
std::vector<std::pair<int, int> > B[3], A[8][3];
inline bool chk(int x) {
  int p[3] = {0, 1, 2};
  do {
    for (int o = 0; o < 4; o++) {
      int u = A[o << 1][p[0]][x - 1].first + 1;
      int y = x, v = ((int)1e9 + 1);
      for (auto [i, j] : A[o << 1][p[1]])
        if (i >= u && !--y) {
          v = i + 1;
          break;
        }
      y = x;
      for (auto [i, j] : A[o << 1][p[2]])
        if (i >= v && !--y) return 1;
      y = x, v = ((int)1e9 + 1);
      for (auto [i, j] : A[o << 1 | 1][p[1]])
        if (i >= u && !--y) {
          v = j + 1;
          break;
        }
      y = x;
      for (auto [i, j] : A[o << 1 | 1][p[2]])
        if (i >= u && j >= v && !--y) return 1;
    }
  } while (std::next_permutation(p, p + 3));
  return 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x, y, c;
    scanf("%d%d%d", &x, &y, &c);
    B[c - 1].emplace_back(x, y);
  }
  for (int oo = 0; oo < 2; oo++) {
    for (int o = 0; o < 2; o++) {
      for (int i = 0; i < 3; i++) {
        A[(oo << 2) | (o << 1)][i] = B[i];
        std::sort(A[(oo << 2) | (o << 1)][i].begin(),
                  A[(oo << 2) | (o << 1)][i].end());
        A[(oo << 2) | (o << 1) | 1][i] = B[i];
        std::sort(A[(oo << 2) | (o << 1) | 1][i].begin(),
                  A[(oo << 2) | (o << 1) | 1][i].end(),
                  [](std::pair<int, int> u, std::pair<int, int> v) {
                    return u.second == v.second ? u.first < v.first
                                                : u.second < v.second;
                  });
      }
      for (int i = 0; i < 3; i++)
        for (int j = 0; j < n / 3; j++) B[i][j].first *= -1;
    }
    for (int i = 0; i < 3; i++)
      for (int j = 0; j < n / 3; j++) std::swap(B[i][j].first, B[i][j].second);
  }
  int L = 1, R = n / 3, res = L;
  while (L <= R) {
    int mid = (L + R) >> 1;
    if (chk(mid))
      res = mid, L = mid + 1;
    else
      R = mid - 1;
  }
  printf("%d\n", 3 * res);
}

```