#include <bits/stdc++.h>
using namespace std;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long max_n = 100005;
long long ptr[3];
pair<long long, long long> points[3][max_n], test_points[3][max_n];
long long x[3][max_n], y[3][max_n];
long long n;
bool check_config(long long cnt, bool flip_x, bool flip_y, bool flip_xy) {
  for (long long i = 1; i <= n / 3; i++) {
    for (long long c = 0; c < 3; c++) test_points[c][i] = points[c][i];
  }
  for (long long i = 1; i <= n / 3; i++) {
    for (long long c = 0; c < 3; c++) {
      if (flip_xy) swap(test_points[c][i].first, test_points[c][i].second);
      if (flip_x) test_points[c][i].first = -test_points[c][i].first;
      if (flip_y) test_points[c][i].second = -test_points[c][i].second;
    }
  }
  for (long long i = 1; i <= n / 3; i++) {
    for (long long c = 0; c < 3; c++) x[c][i] = test_points[c][i].first;
  }
  vector<long long> color_order = {0, 1, 2};
  do {
    nth_element(x[color_order[0]] + 1, x[color_order[0]] + cnt,
                x[color_order[0]] + n / 3 + 1);
    long long bound_0 = x[color_order[0]][cnt];
    long long lesser_1 = 0;
    for (long long i = 1; i <= n / 3; i++) {
      if (x[color_order[1]][i] <= bound_0) lesser_1++;
    }
    if (lesser_1 + cnt > n / 3) continue;
    nth_element(x[color_order[1]] + 1, x[color_order[1]] + lesser_1 + cnt,
                x[color_order[1]] + n / 3 + 1);
    long long bound_1 = x[color_order[1]][lesser_1 + cnt];
    long long lesser_2 = 0;
    for (long long i = 1; i <= n / 3; i++) {
      if (x[color_order[2]][i] <= bound_1) lesser_2++;
    }
    if (lesser_2 + cnt <= n / 3) return 1;
  } while (next_permutation(color_order.begin(), color_order.end()));
  color_order = {0, 1, 2};
  do {
    nth_element(x[color_order[0]] + 1, x[color_order[0]] + cnt,
                x[color_order[0]] + n / 3 + 1);
    long long bound_0 = x[color_order[0]][cnt];
    long long lesser_1 = 0;
    for (long long i = 1; i <= n / 3; i++) {
      if (x[color_order[1]][i] <= bound_0) lesser_1++;
    }
    if (lesser_1 + cnt > n / 3) continue;
    long long cnt_greater = 0;
    for (long long i = 1; i <= n / 3; i++) {
      if (test_points[color_order[1]][i].first > bound_0) {
        y[color_order[1]][cnt_greater + 1] =
            test_points[color_order[1]][i].second;
        cnt_greater++;
      }
    }
    nth_element(y[color_order[1]] + 1, y[color_order[1]] + cnt,
                y[color_order[1]] + cnt_greater + 1);
    long long bound_1 = y[color_order[1]][cnt];
    long long sat = 0;
    for (long long i = 1; i <= n / 3; i++) {
      if (test_points[color_order[2]][i].first > bound_0 &&
          test_points[color_order[2]][i].second > bound_1)
        sat++;
    }
    if (sat >= cnt) return 1;
  } while (next_permutation(color_order.begin(), color_order.end()));
  return 0;
}
bool check(long long cnt) {
  for (long long flip_x = 0; flip_x <= 1; flip_x++) {
    for (long long flip_y = 0; flip_y <= 1; flip_y++) {
      for (long long flip_xy = 0; flip_xy <= 1; flip_xy++) {
        if (check_config(cnt, flip_x, flip_y, flip_xy)) {
          return 1;
        }
      }
    }
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    long long x_i, y_i, c_i;
    cin >> x_i >> y_i >> c_i;
    points[c_i - 1][ptr[c_i - 1] + 1] = make_pair(x_i, y_i);
    ptr[c_i - 1]++;
  }
  long long l = 1, r = n / 3;
  while (l < r) {
    long long mid = (l + r + 1) / 2;
    if (check(mid))
      l = mid;
    else
      r = mid - 1;
  }
  cout << 3 * l << "\n";
}

### Prompt

Construct a cpp code solution to the problem outlined:

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_64 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long max_n = 100005;
long long ptr[3];
pair<long long, long long> points[3][max_n], test_points[3][max_n];
long long x[3][max_n], y[3][max_n];
long long n;
bool check_config(long long cnt, bool flip_x, bool flip_y, bool flip_xy) {
  for (long long i = 1; i <= n / 3; i++) {
    for (long long c = 0; c < 3; c++) test_points[c][i] = points[c][i];
  }
  for (long long i = 1; i <= n / 3; i++) {
    for (long long c = 0; c < 3; c++) {
      if (flip_xy) swap(test_points[c][i].first, test_points[c][i].second);
      if (flip_x) test_points[c][i].first = -test_points[c][i].first;
      if (flip_y) test_points[c][i].second = -test_points[c][i].second;
    }
  }
  for (long long i = 1; i <= n / 3; i++) {
    for (long long c = 0; c < 3; c++) x[c][i] = test_points[c][i].first;
  }
  vector<long long> color_order = {0, 1, 2};
  do {
    nth_element(x[color_order[0]] + 1, x[color_order[0]] + cnt,
                x[color_order[0]] + n / 3 + 1);
    long long bound_0 = x[color_order[0]][cnt];
    long long lesser_1 = 0;
    for (long long i = 1; i <= n / 3; i++) {
      if (x[color_order[1]][i] <= bound_0) lesser_1++;
    }
    if (lesser_1 + cnt > n / 3) continue;
    nth_element(x[color_order[1]] + 1, x[color_order[1]] + lesser_1 + cnt,
                x[color_order[1]] + n / 3 + 1);
    long long bound_1 = x[color_order[1]][lesser_1 + cnt];
    long long lesser_2 = 0;
    for (long long i = 1; i <= n / 3; i++) {
      if (x[color_order[2]][i] <= bound_1) lesser_2++;
    }
    if (lesser_2 + cnt <= n / 3) return 1;
  } while (next_permutation(color_order.begin(), color_order.end()));
  color_order = {0, 1, 2};
  do {
    nth_element(x[color_order[0]] + 1, x[color_order[0]] + cnt,
                x[color_order[0]] + n / 3 + 1);
    long long bound_0 = x[color_order[0]][cnt];
    long long lesser_1 = 0;
    for (long long i = 1; i <= n / 3; i++) {
      if (x[color_order[1]][i] <= bound_0) lesser_1++;
    }
    if (lesser_1 + cnt > n / 3) continue;
    long long cnt_greater = 0;
    for (long long i = 1; i <= n / 3; i++) {
      if (test_points[color_order[1]][i].first > bound_0) {
        y[color_order[1]][cnt_greater + 1] =
            test_points[color_order[1]][i].second;
        cnt_greater++;
      }
    }
    nth_element(y[color_order[1]] + 1, y[color_order[1]] + cnt,
                y[color_order[1]] + cnt_greater + 1);
    long long bound_1 = y[color_order[1]][cnt];
    long long sat = 0;
    for (long long i = 1; i <= n / 3; i++) {
      if (test_points[color_order[2]][i].first > bound_0 &&
          test_points[color_order[2]][i].second > bound_1)
        sat++;
    }
    if (sat >= cnt) return 1;
  } while (next_permutation(color_order.begin(), color_order.end()));
  return 0;
}
bool check(long long cnt) {
  for (long long flip_x = 0; flip_x <= 1; flip_x++) {
    for (long long flip_y = 0; flip_y <= 1; flip_y++) {
      for (long long flip_xy = 0; flip_xy <= 1; flip_xy++) {
        if (check_config(cnt, flip_x, flip_y, flip_xy)) {
          return 1;
        }
      }
    }
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    long long x_i, y_i, c_i;
    cin >> x_i >> y_i >> c_i;
    points[c_i - 1][ptr[c_i - 1] + 1] = make_pair(x_i, y_i);
    ptr[c_i - 1]++;
  }
  long long l = 1, r = n / 3;
  while (l < r) {
    long long mid = (l + r + 1) / 2;
    if (check(mid))
      l = mid;
    else
      r = mid - 1;
  }
  cout << 3 * l << "\n";
}

```