#include <bits/stdc++.h>
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
using namespace std;
const long long infll = (long long)1e18 + 228;
const int inf = 1e9 + 228;
const int mod = 998244353;
const int MOD = 1e9 + 7;
using mem = vector<pair<int, int>>;
int solve(mem a, mem b, mem c) {
  sort((a).begin(), (a).end());
  sort((b).begin(), (b).end());
  sort((c).begin(), (c).end());
  int n = a.size();
  int L = 0, R = a.size() + 1;
  while (R - L > 1) {
    bool ok = 0;
    int M = (L + R) / 2;
    int p1 = M - 1;
    int p2 = inf;
    for (int i = 0; i < n; ++i) {
      if (b[i].first > a[p1].first) {
        p2 = i;
        break;
      }
    }
    if (n - p2 >= M) {
      int p3 = inf;
      p2 += M - 1;
      for (int i = 0; i < n; ++i) {
        if (c[i].first > a[p1].first && c[i].first > b[p2].first) {
          p3 = i;
          break;
        }
      }
      ok |= (n - p3 >= M);
    }
    vector<int> yb, yc;
    for (int i = 0; i < n; ++i) {
      if (b[i].first > a[p1].first) {
        yb.emplace_back(b[i].second);
      }
      if (c[i].first > a[p1].first) {
        yc.emplace_back(c[i].second);
      }
    }
    if (yb.size() >= M) {
      nth_element(yb.begin(), yb.begin() + M - 1, yb.end());
      int ct = 0;
      for (auto x : yc) {
        if (x > yb[M - 1]) {
          ++ct;
        }
      }
      ok |= (ct >= M);
    }
    if (ok) {
      L = M;
    } else {
      R = M;
    }
  }
  return L;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  mem a, b, c;
  vector<int> diff;
  for (int i = 0; i < n; ++i) {
    int x, y, col;
    cin >> x >> y >> col;
    if (col == 1) {
      a.emplace_back(x, y);
    } else if (col == 2) {
      b.emplace_back(x, y);
    } else if (col == 3) {
      c.emplace_back(x, y);
    }
  }
  int ans = 0;
  for (int k = 0; k < 4; ++k) {
    ans = max(ans, solve(a, b, c));
    ans = max(ans, solve(a, c, b));
    ans = max(ans, solve(b, a, c));
    ans = max(ans, solve(b, c, a));
    ans = max(ans, solve(c, a, b));
    ans = max(ans, solve(c, b, a));
    for (auto &[x, y] : a) {
      tie(x, y) = make_pair(-y, x);
    }
    for (auto &[x, y] : b) {
      tie(x, y) = make_pair(-y, x);
    }
    for (auto &[x, y] : c) {
      tie(x, y) = make_pair(-y, x);
    }
  }
  cout << 3 * ans << '\n';
}

### 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>
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
using namespace std;
const long long infll = (long long)1e18 + 228;
const int inf = 1e9 + 228;
const int mod = 998244353;
const int MOD = 1e9 + 7;
using mem = vector<pair<int, int>>;
int solve(mem a, mem b, mem c) {
  sort((a).begin(), (a).end());
  sort((b).begin(), (b).end());
  sort((c).begin(), (c).end());
  int n = a.size();
  int L = 0, R = a.size() + 1;
  while (R - L > 1) {
    bool ok = 0;
    int M = (L + R) / 2;
    int p1 = M - 1;
    int p2 = inf;
    for (int i = 0; i < n; ++i) {
      if (b[i].first > a[p1].first) {
        p2 = i;
        break;
      }
    }
    if (n - p2 >= M) {
      int p3 = inf;
      p2 += M - 1;
      for (int i = 0; i < n; ++i) {
        if (c[i].first > a[p1].first && c[i].first > b[p2].first) {
          p3 = i;
          break;
        }
      }
      ok |= (n - p3 >= M);
    }
    vector<int> yb, yc;
    for (int i = 0; i < n; ++i) {
      if (b[i].first > a[p1].first) {
        yb.emplace_back(b[i].second);
      }
      if (c[i].first > a[p1].first) {
        yc.emplace_back(c[i].second);
      }
    }
    if (yb.size() >= M) {
      nth_element(yb.begin(), yb.begin() + M - 1, yb.end());
      int ct = 0;
      for (auto x : yc) {
        if (x > yb[M - 1]) {
          ++ct;
        }
      }
      ok |= (ct >= M);
    }
    if (ok) {
      L = M;
    } else {
      R = M;
    }
  }
  return L;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  mem a, b, c;
  vector<int> diff;
  for (int i = 0; i < n; ++i) {
    int x, y, col;
    cin >> x >> y >> col;
    if (col == 1) {
      a.emplace_back(x, y);
    } else if (col == 2) {
      b.emplace_back(x, y);
    } else if (col == 3) {
      c.emplace_back(x, y);
    }
  }
  int ans = 0;
  for (int k = 0; k < 4; ++k) {
    ans = max(ans, solve(a, b, c));
    ans = max(ans, solve(a, c, b));
    ans = max(ans, solve(b, a, c));
    ans = max(ans, solve(b, c, a));
    ans = max(ans, solve(c, a, b));
    ans = max(ans, solve(c, b, a));
    for (auto &[x, y] : a) {
      tie(x, y) = make_pair(-y, x);
    }
    for (auto &[x, y] : b) {
      tie(x, y) = make_pair(-y, x);
    }
    for (auto &[x, y] : c) {
      tie(x, y) = make_pair(-y, x);
    }
  }
  cout << 3 * ans << '\n';
}

```