#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int n, a[N], b[N], c[N], ans, t[3], m, id[3], lim;
bool vis[3];
struct Node {
  int x, y;
  bool operator<(const Node tmp) const { return x < tmp.x; }
} p[3][N], q[3][N];
bool try_(int l, int r) {
  int cnt = 0, pos, x = id[1];
  for (int i = 1; i <= m; i++) {
    if (q[x][i].y > r || q[x][i].y < l) {
      continue;
    }
    cnt++;
    if (cnt == lim) {
      pos = q[x][i].x;
      break;
    }
  }
  if (cnt < lim) {
    return 0;
  }
  cnt = 0, x = id[2];
  for (int i = 1; i <= m; i++) {
    if (q[x][i].y > r || q[x][i].y < l || q[x][i].x <= pos) {
      continue;
    }
    cnt++;
    if (cnt >= lim) {
      return 1;
    }
  }
  return 0;
}
bool work_() {
  if (try_(-1e9, p[id[0]][m - lim + 1].x - 1)) {
    return 1;
  }
  if (try_(p[id[0]][lim].x + 1, 1e9)) {
    return 1;
  }
  int pos = -1e9 - 1;
  for (int i = 0; i < 3; i++) {
    int u = id[i], cnt = 0;
    for (int j = 1; j <= m; j++) {
      if (p[u][j].x <= pos) {
        continue;
      }
      cnt++;
      if (cnt == lim) {
        pos = p[u][j].x;
        break;
      }
    }
    if (cnt < lim) {
      return 0;
    }
  }
  return 1;
}
bool dfs_(int x) {
  if (x > 2) {
    return work_();
  }
  bool ok = 0;
  for (int i = 0; i < 3; i++) {
    if (!vis[i]) {
      vis[i] = 1, id[x] = i;
      ok = dfs_(x + 1);
      vis[i] = 0;
      if (ok) {
        break;
      }
    }
  }
  return ok;
}
bool check_() {
  memset(vis, 0, sizeof(vis));
  return dfs_(0);
}
void solve_() {
  t[0] = t[1] = t[2] = 0, m = n / 3;
  for (int i = 1; i <= n; i++) {
    t[c[i]]++;
    p[c[i]][t[c[i]]] = (Node){a[i], b[i]};
    q[c[i]][t[c[i]]] = (Node){b[i], a[i]};
  }
  for (int i = 0; i < 3; i++) {
    sort(p[i] + 1, p[i] + m + 1);
    sort(q[i] + 1, q[i] + m + 1);
  }
  int l = 1, r = m;
  while (l < r) {
    int mid = (l + r + 1) >> 1;
    lim = mid;
    if (check_()) {
      l = mid;
    } else {
      r = mid - 1;
    }
  }
  ans = max(ans, l);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d%d", &a[i], &b[i], &c[i]);
    c[i]--;
  }
  solve_();
  for (int i = 1; i <= n; i++) {
    swap(a[i], b[i]);
  }
  solve_();
  printf("%d\n", ans * 3);
}

### 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;
const int N = 1e5 + 10;
int n, a[N], b[N], c[N], ans, t[3], m, id[3], lim;
bool vis[3];
struct Node {
  int x, y;
  bool operator<(const Node tmp) const { return x < tmp.x; }
} p[3][N], q[3][N];
bool try_(int l, int r) {
  int cnt = 0, pos, x = id[1];
  for (int i = 1; i <= m; i++) {
    if (q[x][i].y > r || q[x][i].y < l) {
      continue;
    }
    cnt++;
    if (cnt == lim) {
      pos = q[x][i].x;
      break;
    }
  }
  if (cnt < lim) {
    return 0;
  }
  cnt = 0, x = id[2];
  for (int i = 1; i <= m; i++) {
    if (q[x][i].y > r || q[x][i].y < l || q[x][i].x <= pos) {
      continue;
    }
    cnt++;
    if (cnt >= lim) {
      return 1;
    }
  }
  return 0;
}
bool work_() {
  if (try_(-1e9, p[id[0]][m - lim + 1].x - 1)) {
    return 1;
  }
  if (try_(p[id[0]][lim].x + 1, 1e9)) {
    return 1;
  }
  int pos = -1e9 - 1;
  for (int i = 0; i < 3; i++) {
    int u = id[i], cnt = 0;
    for (int j = 1; j <= m; j++) {
      if (p[u][j].x <= pos) {
        continue;
      }
      cnt++;
      if (cnt == lim) {
        pos = p[u][j].x;
        break;
      }
    }
    if (cnt < lim) {
      return 0;
    }
  }
  return 1;
}
bool dfs_(int x) {
  if (x > 2) {
    return work_();
  }
  bool ok = 0;
  for (int i = 0; i < 3; i++) {
    if (!vis[i]) {
      vis[i] = 1, id[x] = i;
      ok = dfs_(x + 1);
      vis[i] = 0;
      if (ok) {
        break;
      }
    }
  }
  return ok;
}
bool check_() {
  memset(vis, 0, sizeof(vis));
  return dfs_(0);
}
void solve_() {
  t[0] = t[1] = t[2] = 0, m = n / 3;
  for (int i = 1; i <= n; i++) {
    t[c[i]]++;
    p[c[i]][t[c[i]]] = (Node){a[i], b[i]};
    q[c[i]][t[c[i]]] = (Node){b[i], a[i]};
  }
  for (int i = 0; i < 3; i++) {
    sort(p[i] + 1, p[i] + m + 1);
    sort(q[i] + 1, q[i] + m + 1);
  }
  int l = 1, r = m;
  while (l < r) {
    int mid = (l + r + 1) >> 1;
    lim = mid;
    if (check_()) {
      l = mid;
    } else {
      r = mid - 1;
    }
  }
  ans = max(ans, l);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d%d", &a[i], &b[i], &c[i]);
    c[i]--;
  }
  solve_();
  for (int i = 1; i <= n; i++) {
    swap(a[i], b[i]);
  }
  solve_();
  printf("%d\n", ans * 3);
}

```