#include <bits/stdc++.h>
using namespace std;
const int MAXN = (int)1e5 + 5;
struct Point {
  int x, y, c;
  Point() { x = y = c = 0; }
  void rotate() {
    swap(x, y);
    x = -x;
  }
} P[MAXN];
pair<Point, int> Q[MAXN];
Point T[MAXN];
bool operator<(Point a, Point b) {
  return make_pair(a.x, a.y) < make_pair(b.x, b.y);
}
bool byY(Point a, Point b) { return make_pair(a.y, a.x) < make_pair(b.y, b.x); }
int perm[3];
int n;
bool check(int k) {
  iota(perm, perm + 3, 1);
  do {
    int p = 0, cnt = 0;
    while (p < n && (cnt < k || P[p].x == P[p + 1].x)) {
      p++;
      cnt += (P[p].c == perm[0]);
    }
    int q = p;
    cnt = 0;
    while (q < n && (cnt < k || P[q].x == P[q + 1].x)) {
      q++;
      cnt += (P[q].c == perm[1]);
    }
    cnt = 0;
    for (int i = q + 1; i <= n; i++) {
      cnt += (P[i].c == perm[2]);
    }
    if (cnt >= k) {
      return 1;
    }
    int m = 0;
    for (int i = 1; i <= n; i++) {
      if (Q[i].second > p) {
        T[++m] = Q[i].first;
      }
    }
    q = 0;
    cnt = 0;
    while (q < m && (cnt < k || T[q].y == T[q + 1].y)) {
      q++;
      cnt += (T[q].c == perm[1]);
    }
    cnt = 0;
    for (int i = q + 1; i <= m; i++) {
      cnt += (T[i].c == perm[2]);
    }
    if (cnt >= k) {
      return 1;
    }
  } while (next_permutation(perm, perm + 3));
  return 0;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> P[i].x >> P[i].y >> P[i].c;
  }
  int ans = 0;
  for (int i = 0; i < 3; i++) {
    perm[i] = i;
  }
  for (int rot = 0; rot < 4; rot++) {
    sort(P + 1, P + n + 1);
    for (int i = 1; i <= n; i++) {
      Q[i].first = P[i];
      Q[i].second = i;
    }
    sort(Q + 1, Q + n + 1, [&](pair<Point, int> a, pair<Point, int> b) -> bool {
      return make_pair(a.first.y, a.first.x) < make_pair(b.first.y, b.first.x);
    });
    int l = 1, r = n / 3, opt = 0;
    while (l <= r) {
      int mid = (l + r) / 2;
      if (check(mid)) {
        opt = mid;
        l = mid + 1;
      } else {
        r = mid - 1;
      }
    }
    ans = max(ans, opt);
    for (int i = 1; i <= n; i++) {
      P[i].rotate();
    }
  }
  cout << ans * 3 << '\n';
  return 0;
}

### 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;
const int MAXN = (int)1e5 + 5;
struct Point {
  int x, y, c;
  Point() { x = y = c = 0; }
  void rotate() {
    swap(x, y);
    x = -x;
  }
} P[MAXN];
pair<Point, int> Q[MAXN];
Point T[MAXN];
bool operator<(Point a, Point b) {
  return make_pair(a.x, a.y) < make_pair(b.x, b.y);
}
bool byY(Point a, Point b) { return make_pair(a.y, a.x) < make_pair(b.y, b.x); }
int perm[3];
int n;
bool check(int k) {
  iota(perm, perm + 3, 1);
  do {
    int p = 0, cnt = 0;
    while (p < n && (cnt < k || P[p].x == P[p + 1].x)) {
      p++;
      cnt += (P[p].c == perm[0]);
    }
    int q = p;
    cnt = 0;
    while (q < n && (cnt < k || P[q].x == P[q + 1].x)) {
      q++;
      cnt += (P[q].c == perm[1]);
    }
    cnt = 0;
    for (int i = q + 1; i <= n; i++) {
      cnt += (P[i].c == perm[2]);
    }
    if (cnt >= k) {
      return 1;
    }
    int m = 0;
    for (int i = 1; i <= n; i++) {
      if (Q[i].second > p) {
        T[++m] = Q[i].first;
      }
    }
    q = 0;
    cnt = 0;
    while (q < m && (cnt < k || T[q].y == T[q + 1].y)) {
      q++;
      cnt += (T[q].c == perm[1]);
    }
    cnt = 0;
    for (int i = q + 1; i <= m; i++) {
      cnt += (T[i].c == perm[2]);
    }
    if (cnt >= k) {
      return 1;
    }
  } while (next_permutation(perm, perm + 3));
  return 0;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> P[i].x >> P[i].y >> P[i].c;
  }
  int ans = 0;
  for (int i = 0; i < 3; i++) {
    perm[i] = i;
  }
  for (int rot = 0; rot < 4; rot++) {
    sort(P + 1, P + n + 1);
    for (int i = 1; i <= n; i++) {
      Q[i].first = P[i];
      Q[i].second = i;
    }
    sort(Q + 1, Q + n + 1, [&](pair<Point, int> a, pair<Point, int> b) -> bool {
      return make_pair(a.first.y, a.first.x) < make_pair(b.first.y, b.first.x);
    });
    int l = 1, r = n / 3, opt = 0;
    while (l <= r) {
      int mid = (l + r) / 2;
      if (check(mid)) {
        opt = mid;
        l = mid + 1;
      } else {
        r = mid - 1;
      }
    }
    ans = max(ans, opt);
    for (int i = 1; i <= n; i++) {
      P[i].rotate();
    }
  }
  cout << ans * 3 << '\n';
  return 0;
}

```