#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5, inf = 1e9;
struct P {
  int x, y, c;
} p[N];
int n, ans, R[5];
void sol1(vector<int>& v) {
  int l = 1, r = n / 3, f = 0;
  while (l <= r) {
    int mid = (l + r) >> 1, j = 0, k = 0;
    for (int i = 0; i < n && j < 3; i++)
      if (v[j] == p[i].c && ++k == mid) ++j, k = 0;
    if (j == 3)
      l = mid + 1, f = mid;
    else
      r = mid - 1;
  }
  ans = max(ans, 3 * f);
}
void sol2(int q) {
  int a, b;
  if (q == 1) a = 2, b = 3;
  if (q == 2) a = 1, b = 3;
  if (q == 3) a = 1, b = 2;
  int l = 1, r = n / 3, f = 0;
  while (l <= r) {
    int mid = (l + r) >> 1, s = n, k = 0;
    for (int i = 0; i < n; i++)
      if (p[i].c == q && ++k == mid) {
        s = i + 1;
        break;
      }
    vector<P> v;
    for (int i = s; i < n; i++)
      if (p[i].c != q) v.push_back(p[i]);
    sort(v.begin(), v.end(), [&](P a, P b) { return a.y < b.y; });
    int la = inf, ra = -inf, lb = inf, rb = -inf, t = 0;
    for (auto& it : v)
      if (it.c == a && ++t == mid) {
        la = it.y;
        break;
      }
    t = 0;
    for (auto& it : v)
      if (it.c == b && ++t == mid) {
        lb = it.y;
        break;
      }
    reverse(v.begin(), v.end());
    t = 0;
    for (auto& it : v)
      if (it.c == a && ++t == mid) {
        ra = it.y;
        break;
      }
    t = 0;
    for (auto& it : v)
      if (it.c == b && ++t == mid) {
        rb = it.y;
        break;
      }
    if (la < rb || lb < ra)
      l = mid + 1, f = mid;
    else
      r = mid - 1;
  }
  ans = max(ans, 3 * f);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d%d%d", &p[i].x, &p[i].y, &p[i].c);
  for (int i = 0; i < 4; i++) {
    if (i < 2) {
      vector<int> v = {1, 2, 3};
      do {
        for (int i = 0; i < 3; i++) R[v[i]] = i;
        sort(p, p + n, [&](P a, P b) {
          if (a.x == b.x) return R[a.c] > R[b.c];
          return a.x < b.x;
        });
        sol1(v);
      } while (next_permutation(v.begin(), v.end()));
    }
    for (int j = 1; j <= 3; j++) {
      sort(p, p + n, [&](P a, P b) {
        if (a.x == b.x) return (a.c == j) < (b.c == j);
        return a.x < b.x;
      });
      sol2(j);
    }
    for (int j = 0; j < n; j++) p[j] = {p[j].y, n - p[j].x - 1, p[j].c};
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Create a solution in cpp for the following problem:

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 + 5, inf = 1e9;
struct P {
  int x, y, c;
} p[N];
int n, ans, R[5];
void sol1(vector<int>& v) {
  int l = 1, r = n / 3, f = 0;
  while (l <= r) {
    int mid = (l + r) >> 1, j = 0, k = 0;
    for (int i = 0; i < n && j < 3; i++)
      if (v[j] == p[i].c && ++k == mid) ++j, k = 0;
    if (j == 3)
      l = mid + 1, f = mid;
    else
      r = mid - 1;
  }
  ans = max(ans, 3 * f);
}
void sol2(int q) {
  int a, b;
  if (q == 1) a = 2, b = 3;
  if (q == 2) a = 1, b = 3;
  if (q == 3) a = 1, b = 2;
  int l = 1, r = n / 3, f = 0;
  while (l <= r) {
    int mid = (l + r) >> 1, s = n, k = 0;
    for (int i = 0; i < n; i++)
      if (p[i].c == q && ++k == mid) {
        s = i + 1;
        break;
      }
    vector<P> v;
    for (int i = s; i < n; i++)
      if (p[i].c != q) v.push_back(p[i]);
    sort(v.begin(), v.end(), [&](P a, P b) { return a.y < b.y; });
    int la = inf, ra = -inf, lb = inf, rb = -inf, t = 0;
    for (auto& it : v)
      if (it.c == a && ++t == mid) {
        la = it.y;
        break;
      }
    t = 0;
    for (auto& it : v)
      if (it.c == b && ++t == mid) {
        lb = it.y;
        break;
      }
    reverse(v.begin(), v.end());
    t = 0;
    for (auto& it : v)
      if (it.c == a && ++t == mid) {
        ra = it.y;
        break;
      }
    t = 0;
    for (auto& it : v)
      if (it.c == b && ++t == mid) {
        rb = it.y;
        break;
      }
    if (la < rb || lb < ra)
      l = mid + 1, f = mid;
    else
      r = mid - 1;
  }
  ans = max(ans, 3 * f);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d%d%d", &p[i].x, &p[i].y, &p[i].c);
  for (int i = 0; i < 4; i++) {
    if (i < 2) {
      vector<int> v = {1, 2, 3};
      do {
        for (int i = 0; i < 3; i++) R[v[i]] = i;
        sort(p, p + n, [&](P a, P b) {
          if (a.x == b.x) return R[a.c] > R[b.c];
          return a.x < b.x;
        });
        sol1(v);
      } while (next_permutation(v.begin(), v.end()));
    }
    for (int j = 1; j <= 3; j++) {
      sort(p, p + n, [&](P a, P b) {
        if (a.x == b.x) return (a.c == j) < (b.c == j);
        return a.x < b.x;
      });
      sol2(j);
    }
    for (int j = 0; j < n; j++) p[j] = {p[j].y, n - p[j].x - 1, p[j].c};
  }
  printf("%d\n", ans);
  return 0;
}

```