#include <bits/stdc++.h>
using namespace std;
int n, ans, bb, t[10], id[10], l, tot, lim;
struct item {
  int x, y, c;
} nx[110000], ny[110000];
bool cmpx(item a, item b) { return a.x < b.x; }
bool cmpy(item a, item b) { return a.y < b.y; }
int scan(int c, int k) {
  tot = 0;
  for (; l <= n; l++) {
    if (nx[l].y <= lim) continue;
    if (nx[l].c == c) {
      tot++;
      if (tot == k) break;
    }
  }
  int f = 0;
  for (int i = l; i <= n; i++) {
    if (nx[i].x != nx[l].x) {
      l = i;
      f = 1;
      break;
    }
  }
  if (f == 0) l = n + 1;
  if (tot != k) return 0;
  return 1;
}
int check1(int k) {
  l = 1;
  if (!scan(t[1], k)) return 0;
  if (!scan(t[2], k)) return 0;
  if (!scan(t[3], k)) return 0;
  return 1;
}
void binary1(int l, int r) {
  if (l > r) return;
  int mid = l + r >> 1;
  if (check1(mid))
    bb = mid, binary1(mid + 1, r);
  else
    binary1(l, mid - 1);
}
int check2(int k) {
  l = 1, tot = 0;
  for (int i = 1; i <= n; i++) {
    if (ny[i].c == t[1]) {
      tot++;
      lim = ny[i].y;
      if (tot == k) break;
    }
  }
  if (!scan(t[2], k)) return 0;
  if (!scan(t[3], k)) return 0;
  return 1;
}
void binary2(int l, int r) {
  if (l > r) return;
  int mid = l + r >> 1;
  if (check2(mid))
    bb = mid, binary2(mid + 1, r);
  else
    binary2(l, mid - 1);
}
int main() {
  scanf("%d", &n);
  lim = -2e9;
  for (int i = 1; i <= n; i++) scanf("%d%d%d", &nx[i].x, &nx[i].y, &nx[i].c);
  sort(nx + 1, nx + n + 1, cmpx);
  t[1] = 1;
  t[2] = 2;
  t[3] = 3;
  for (int i = 1; i <= 6; i++) {
    binary1(1, n / 3);
    ans = max(ans, bb);
    if (i != 6) next_permutation(t + 1, t + 4);
  }
  t[1] = 1;
  t[2] = 2;
  t[3] = 3;
  for (int i = 1; i <= n; i++) swap(nx[i].x, nx[i].y);
  sort(nx + 1, nx + n + 1, cmpx);
  for (int i = 1; i <= 6; i++) {
    binary1(1, n / 3);
    ans = max(ans, bb);
    if (i != 6) next_permutation(t + 1, t + 4);
  }
  t[1] = 1;
  t[2] = 2;
  t[3] = 3;
  for (int i = 1; i <= n; i++) swap(nx[i].x, nx[i].y);
  memcpy(ny, nx, sizeof(ny));
  sort(nx + 1, nx + n + 1, cmpx);
  sort(ny + 1, ny + n + 1, cmpy);
  for (int i = 1; i <= 6; i++) {
    binary2(1, n / 3);
    ans = max(ans, bb);
    if (i != 6) next_permutation(t + 1, t + 4);
  }
  t[1] = 1;
  t[2] = 2;
  t[3] = 3;
  for (int i = 1; i <= n; i++) nx[i].y = -nx[i].y;
  memcpy(ny, nx, sizeof(ny));
  sort(nx + 1, nx + n + 1, cmpx);
  sort(ny + 1, ny + n + 1, cmpy);
  for (int i = 1; i <= 6; i++) {
    binary2(1, n / 3);
    ans = max(ans, bb);
    if (i != 6) next_permutation(t + 1, t + 4);
  }
  t[1] = 1;
  t[2] = 2;
  t[3] = 3;
  for (int i = 1; i <= n; i++) swap(nx[i].x, nx[i].y);
  memcpy(ny, nx, sizeof(ny));
  sort(nx + 1, nx + n + 1, cmpx);
  sort(ny + 1, ny + n + 1, cmpy);
  for (int i = 1; i <= 6; i++) {
    binary2(1, n / 3);
    ans = max(ans, bb);
    if (i != 6) next_permutation(t + 1, t + 4);
  }
  t[1] = 1;
  t[2] = 2;
  t[3] = 3;
  for (int i = 1; i <= n; i++) nx[i].y = -nx[i].y;
  memcpy(ny, nx, sizeof(ny));
  sort(nx + 1, nx + n + 1, cmpx);
  sort(ny + 1, ny + n + 1, cmpy);
  for (int i = 1; i <= 6; i++) {
    binary2(1, n / 3);
    ans = max(ans, bb);
    if (i != 6) next_permutation(t + 1, t + 4);
  }
  cout << ans * 3;
  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;
int n, ans, bb, t[10], id[10], l, tot, lim;
struct item {
  int x, y, c;
} nx[110000], ny[110000];
bool cmpx(item a, item b) { return a.x < b.x; }
bool cmpy(item a, item b) { return a.y < b.y; }
int scan(int c, int k) {
  tot = 0;
  for (; l <= n; l++) {
    if (nx[l].y <= lim) continue;
    if (nx[l].c == c) {
      tot++;
      if (tot == k) break;
    }
  }
  int f = 0;
  for (int i = l; i <= n; i++) {
    if (nx[i].x != nx[l].x) {
      l = i;
      f = 1;
      break;
    }
  }
  if (f == 0) l = n + 1;
  if (tot != k) return 0;
  return 1;
}
int check1(int k) {
  l = 1;
  if (!scan(t[1], k)) return 0;
  if (!scan(t[2], k)) return 0;
  if (!scan(t[3], k)) return 0;
  return 1;
}
void binary1(int l, int r) {
  if (l > r) return;
  int mid = l + r >> 1;
  if (check1(mid))
    bb = mid, binary1(mid + 1, r);
  else
    binary1(l, mid - 1);
}
int check2(int k) {
  l = 1, tot = 0;
  for (int i = 1; i <= n; i++) {
    if (ny[i].c == t[1]) {
      tot++;
      lim = ny[i].y;
      if (tot == k) break;
    }
  }
  if (!scan(t[2], k)) return 0;
  if (!scan(t[3], k)) return 0;
  return 1;
}
void binary2(int l, int r) {
  if (l > r) return;
  int mid = l + r >> 1;
  if (check2(mid))
    bb = mid, binary2(mid + 1, r);
  else
    binary2(l, mid - 1);
}
int main() {
  scanf("%d", &n);
  lim = -2e9;
  for (int i = 1; i <= n; i++) scanf("%d%d%d", &nx[i].x, &nx[i].y, &nx[i].c);
  sort(nx + 1, nx + n + 1, cmpx);
  t[1] = 1;
  t[2] = 2;
  t[3] = 3;
  for (int i = 1; i <= 6; i++) {
    binary1(1, n / 3);
    ans = max(ans, bb);
    if (i != 6) next_permutation(t + 1, t + 4);
  }
  t[1] = 1;
  t[2] = 2;
  t[3] = 3;
  for (int i = 1; i <= n; i++) swap(nx[i].x, nx[i].y);
  sort(nx + 1, nx + n + 1, cmpx);
  for (int i = 1; i <= 6; i++) {
    binary1(1, n / 3);
    ans = max(ans, bb);
    if (i != 6) next_permutation(t + 1, t + 4);
  }
  t[1] = 1;
  t[2] = 2;
  t[3] = 3;
  for (int i = 1; i <= n; i++) swap(nx[i].x, nx[i].y);
  memcpy(ny, nx, sizeof(ny));
  sort(nx + 1, nx + n + 1, cmpx);
  sort(ny + 1, ny + n + 1, cmpy);
  for (int i = 1; i <= 6; i++) {
    binary2(1, n / 3);
    ans = max(ans, bb);
    if (i != 6) next_permutation(t + 1, t + 4);
  }
  t[1] = 1;
  t[2] = 2;
  t[3] = 3;
  for (int i = 1; i <= n; i++) nx[i].y = -nx[i].y;
  memcpy(ny, nx, sizeof(ny));
  sort(nx + 1, nx + n + 1, cmpx);
  sort(ny + 1, ny + n + 1, cmpy);
  for (int i = 1; i <= 6; i++) {
    binary2(1, n / 3);
    ans = max(ans, bb);
    if (i != 6) next_permutation(t + 1, t + 4);
  }
  t[1] = 1;
  t[2] = 2;
  t[3] = 3;
  for (int i = 1; i <= n; i++) swap(nx[i].x, nx[i].y);
  memcpy(ny, nx, sizeof(ny));
  sort(nx + 1, nx + n + 1, cmpx);
  sort(ny + 1, ny + n + 1, cmpy);
  for (int i = 1; i <= 6; i++) {
    binary2(1, n / 3);
    ans = max(ans, bb);
    if (i != 6) next_permutation(t + 1, t + 4);
  }
  t[1] = 1;
  t[2] = 2;
  t[3] = 3;
  for (int i = 1; i <= n; i++) nx[i].y = -nx[i].y;
  memcpy(ny, nx, sizeof(ny));
  sort(nx + 1, nx + n + 1, cmpx);
  sort(ny + 1, ny + n + 1, cmpy);
  for (int i = 1; i <= 6; i++) {
    binary2(1, n / 3);
    ans = max(ans, bb);
    if (i != 6) next_permutation(t + 1, t + 4);
  }
  cout << ans * 3;
  return 0;
}

```