#include <bits/stdc++.h>
using namespace std;
struct SegmentTree {
  int n;
  vector<int> cnt[2];
  explicit SegmentTree(int n) : n(n) {
    cnt[0].assign(n << 2 | 1, 0);
    cnt[1].assign(n << 2 | 1, 0);
  }
  void Update(int node, int l, int r, int x, int zero, int one) {
    if (x < l || x > r) return;
    if (l == r) {
      cnt[0][node] += zero;
      cnt[1][node] += one;
      return;
    }
    int m = l + r >> 1;
    Update(node << 1, l, m, x, zero, one);
    Update(node << 1 | 1, m + 1, r, x, zero, one);
    cnt[0][node] = cnt[0][node << 1] + cnt[0][node << 1 | 1];
    cnt[1][node] = cnt[1][node << 1] + cnt[1][node << 1 | 1];
  }
  int GetMin(int node, int l, int r, int zero, int one) {
    if (l == r) {
      return min(zero + cnt[0][node], one);
    }
    int m = l + r >> 1;
    int res = 0;
    auto cur_zero = zero + cnt[0][node << 1];
    auto cur_one = one + cnt[1][node << 1 | 1];
    res = max(res, min(cur_zero, cur_one));
    if (cur_zero > cur_one) {
      res = max(res, GetMin(node << 1, l, m, zero, cur_one));
    } else {
      res = max(res, GetMin(node << 1 | 1, m + 1, r, cur_zero, one));
    }
    return res;
  }
};
struct Point {
  int x, y, c;
};
string to_string(Point u) {
  return "{" + to_string(u.x) + ", " + to_string(u.y) + "," + to_string(u.c) +
         "}";
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<Point> a(n);
  vector<int> vals;
  for (auto &u : a) {
    cin >> u.x >> u.y >> u.c;
    vals.push_back(u.x);
    vals.push_back(u.y);
    u.c--;
  }
  auto Normalize = [&](vector<int> &a) {
    sort(a.begin(), a.end());
    a.resize(unique(a.begin(), a.end()) - a.begin());
  };
  Normalize(vals);
  auto m = vals.size();
  for (auto &u : a) {
    u.x = lower_bound(vals.begin(), vals.end(), u.x) - vals.begin();
    u.y = lower_bound(vals.begin(), vals.end(), u.y) - vals.begin();
  }
  int res = 0;
  auto Solve = [&](const vector<Point> &b) {
    int num = n / 3;
    int cur = 0;
    SegmentTree t_x(m);
    SegmentTree t_y(m);
    for (int i = 0; i < m; i++) {
      while (cur < n && b[cur].x <= i) {
        if (b[cur].c == 0) {
          num--;
        } else if (b[cur].c == 1) {
          t_x.Update(1, 0, m - 1, b[cur].x, 1, 0);
          t_y.Update(1, 0, m - 1, b[cur].y, 1, 0);
        } else if (b[cur].c == 2) {
          t_x.Update(1, 0, m - 1, b[cur].x, 0, 1);
          t_y.Update(1, 0, m - 1, b[cur].y, 0, 1);
        } else {
          exit(1);
        }
        cur++;
      }
      res = max(res, min(num, t_x.GetMin(1, 0, m - 1, 0, 0)));
      res = max(res, min(num, t_y.GetMin(1, 0, m - 1, 0, 0)));
    }
  };
  auto SolveAll = [&](const vector<Point> &a) {
    vector<int> id(3);
    iota(id.begin(), id.end(), 0);
    while (1) {
      auto b = a;
      for (auto &p : b) {
        p.c = id[p.c];
      }
      Solve(b);
      if (!next_permutation(id.begin(), id.end())) break;
    }
  };
  for (int z = 0; z < 2; z++) {
    for (auto &p : a) {
      if (z) swap(p.x, p.y);
    }
    sort(a.begin(), a.end(),
         [&](const Point &u, const Point &v) { return u.x < v.x; });
    SolveAll(a);
    for (auto &p : a) {
      p.x = m - p.x - 1;
    }
    reverse(a.begin(), a.end());
    SolveAll(a);
  }
  cout << res * 3 << '\n';
  return 0;
}

### Prompt

Generate a cpp solution to 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;
struct SegmentTree {
  int n;
  vector<int> cnt[2];
  explicit SegmentTree(int n) : n(n) {
    cnt[0].assign(n << 2 | 1, 0);
    cnt[1].assign(n << 2 | 1, 0);
  }
  void Update(int node, int l, int r, int x, int zero, int one) {
    if (x < l || x > r) return;
    if (l == r) {
      cnt[0][node] += zero;
      cnt[1][node] += one;
      return;
    }
    int m = l + r >> 1;
    Update(node << 1, l, m, x, zero, one);
    Update(node << 1 | 1, m + 1, r, x, zero, one);
    cnt[0][node] = cnt[0][node << 1] + cnt[0][node << 1 | 1];
    cnt[1][node] = cnt[1][node << 1] + cnt[1][node << 1 | 1];
  }
  int GetMin(int node, int l, int r, int zero, int one) {
    if (l == r) {
      return min(zero + cnt[0][node], one);
    }
    int m = l + r >> 1;
    int res = 0;
    auto cur_zero = zero + cnt[0][node << 1];
    auto cur_one = one + cnt[1][node << 1 | 1];
    res = max(res, min(cur_zero, cur_one));
    if (cur_zero > cur_one) {
      res = max(res, GetMin(node << 1, l, m, zero, cur_one));
    } else {
      res = max(res, GetMin(node << 1 | 1, m + 1, r, cur_zero, one));
    }
    return res;
  }
};
struct Point {
  int x, y, c;
};
string to_string(Point u) {
  return "{" + to_string(u.x) + ", " + to_string(u.y) + "," + to_string(u.c) +
         "}";
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<Point> a(n);
  vector<int> vals;
  for (auto &u : a) {
    cin >> u.x >> u.y >> u.c;
    vals.push_back(u.x);
    vals.push_back(u.y);
    u.c--;
  }
  auto Normalize = [&](vector<int> &a) {
    sort(a.begin(), a.end());
    a.resize(unique(a.begin(), a.end()) - a.begin());
  };
  Normalize(vals);
  auto m = vals.size();
  for (auto &u : a) {
    u.x = lower_bound(vals.begin(), vals.end(), u.x) - vals.begin();
    u.y = lower_bound(vals.begin(), vals.end(), u.y) - vals.begin();
  }
  int res = 0;
  auto Solve = [&](const vector<Point> &b) {
    int num = n / 3;
    int cur = 0;
    SegmentTree t_x(m);
    SegmentTree t_y(m);
    for (int i = 0; i < m; i++) {
      while (cur < n && b[cur].x <= i) {
        if (b[cur].c == 0) {
          num--;
        } else if (b[cur].c == 1) {
          t_x.Update(1, 0, m - 1, b[cur].x, 1, 0);
          t_y.Update(1, 0, m - 1, b[cur].y, 1, 0);
        } else if (b[cur].c == 2) {
          t_x.Update(1, 0, m - 1, b[cur].x, 0, 1);
          t_y.Update(1, 0, m - 1, b[cur].y, 0, 1);
        } else {
          exit(1);
        }
        cur++;
      }
      res = max(res, min(num, t_x.GetMin(1, 0, m - 1, 0, 0)));
      res = max(res, min(num, t_y.GetMin(1, 0, m - 1, 0, 0)));
    }
  };
  auto SolveAll = [&](const vector<Point> &a) {
    vector<int> id(3);
    iota(id.begin(), id.end(), 0);
    while (1) {
      auto b = a;
      for (auto &p : b) {
        p.c = id[p.c];
      }
      Solve(b);
      if (!next_permutation(id.begin(), id.end())) break;
    }
  };
  for (int z = 0; z < 2; z++) {
    for (auto &p : a) {
      if (z) swap(p.x, p.y);
    }
    sort(a.begin(), a.end(),
         [&](const Point &u, const Point &v) { return u.x < v.x; });
    SolveAll(a);
    for (auto &p : a) {
      p.x = m - p.x - 1;
    }
    reverse(a.begin(), a.end());
    SolveAll(a);
  }
  cout << res * 3 << '\n';
  return 0;
}

```