#include <bits/stdc++.h>
using namespace std;
const int max_n = 100111, inf = 1000111222;
struct segment_tree {
  int sum[4 * max_n][2];
  void clr() { memset(sum, 0, sizeof(sum)); }
  void update(int v, int l, int r, int x, int c) {
    if (l == r) {
      ++sum[v][c];
      return;
    }
    int mid = (l + r) / 2;
    if (x <= mid) {
      update(2 * v, l, mid, x, c);
    } else {
      update(2 * v + 1, mid + 1, r, x, c);
    }
    sum[v][0] = sum[2 * v][0] + sum[2 * v + 1][0];
    sum[v][1] = sum[2 * v][1] + sum[2 * v + 1][1];
  }
  int get(int v, int l, int r, int add0, int add1) {
    if (l == r) {
      return max(min(add0 + sum[v][0], add1), min(add0, add1 + sum[v][1]));
    }
    int mid = (l + r) / 2;
    if (add0 + sum[2 * v][0] <= add1 + sum[2 * v + 1][1]) {
      return get(2 * v + 1, mid + 1, r, add0 + sum[2 * v][0], add1);
    }
    return get(2 * v, l, mid, add0, add1 + sum[2 * v + 1][1]);
  }
};
struct Point {
  int x, y, c;
  bool operator<(const Point &p) const { return x < p.x; }
};
int n;
Point p[max_n];
vector<int> vx, vy;
segment_tree st;
int compress(const vector<int> &v, int x) {
  return lower_bound(v.begin(), v.end(), x) - v.begin();
}
int solve() {
  sort(p, p + n);
  vector<int> perm{0, 1, 2};
  int res = 0;
  do {
    st.clr();
    int c3 = n / 3;
    for (int i = 0; i < n;) {
      int pos = i;
      while (i < n && p[i].x == p[pos].x) {
        if (p[i].c == perm[2]) {
          --c3;
        } else {
          int tp = (p[i].c == perm[1]);
          st.update(1, 0, n - 1, p[i].x, tp);
        }
        ++i;
      }
      res = max(res, min(c3, st.get(1, 0, n - 1, 0, 0)));
    }
    for (int it = 0; it < 2; ++it) {
      st.clr();
      int c3 = n / 3;
      for (int i = 0; i < n;) {
        int pos = i;
        while (i < n && p[i].x == p[pos].x) {
          if (p[i].c == perm[2]) {
            --c3;
          } else {
            int tp = (p[i].c == perm[1]);
            st.update(1, 0, n - 1, p[i].y, tp);
          }
          ++i;
        }
        res = max(res, min(c3, st.get(1, 0, n - 1, 0, 0)));
      }
      for (int i = 0; i < n; ++i) {
        p[i].x = n - 1 - p[i].x;
      }
      reverse(p, p + n);
    }
  } while (next_permutation(perm.begin(), perm.end()));
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> p[i].x >> p[i].y >> p[i].c;
    --p[i].c;
    vx.push_back(p[i].x);
    vy.push_back(p[i].y);
  }
  sort(vx.begin(), vx.end());
  sort(vy.begin(), vy.end());
  vx.erase(unique(vx.begin(), vx.end()), vx.end());
  vy.erase(unique(vy.begin(), vy.end()), vy.end());
  for (int i = 0; i < n; ++i) {
    p[i].x = compress(vx, p[i].x);
    p[i].y = compress(vy, p[i].y);
  }
  int ans = 0;
  for (int it = 0; it < 2; ++it) {
    ans = max(ans, solve());
    for (int i = 0; i < n; ++i) {
      swap(p[i].x, p[i].y);
    }
  }
  cout << ans * 3 << "\n";
  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;
const int max_n = 100111, inf = 1000111222;
struct segment_tree {
  int sum[4 * max_n][2];
  void clr() { memset(sum, 0, sizeof(sum)); }
  void update(int v, int l, int r, int x, int c) {
    if (l == r) {
      ++sum[v][c];
      return;
    }
    int mid = (l + r) / 2;
    if (x <= mid) {
      update(2 * v, l, mid, x, c);
    } else {
      update(2 * v + 1, mid + 1, r, x, c);
    }
    sum[v][0] = sum[2 * v][0] + sum[2 * v + 1][0];
    sum[v][1] = sum[2 * v][1] + sum[2 * v + 1][1];
  }
  int get(int v, int l, int r, int add0, int add1) {
    if (l == r) {
      return max(min(add0 + sum[v][0], add1), min(add0, add1 + sum[v][1]));
    }
    int mid = (l + r) / 2;
    if (add0 + sum[2 * v][0] <= add1 + sum[2 * v + 1][1]) {
      return get(2 * v + 1, mid + 1, r, add0 + sum[2 * v][0], add1);
    }
    return get(2 * v, l, mid, add0, add1 + sum[2 * v + 1][1]);
  }
};
struct Point {
  int x, y, c;
  bool operator<(const Point &p) const { return x < p.x; }
};
int n;
Point p[max_n];
vector<int> vx, vy;
segment_tree st;
int compress(const vector<int> &v, int x) {
  return lower_bound(v.begin(), v.end(), x) - v.begin();
}
int solve() {
  sort(p, p + n);
  vector<int> perm{0, 1, 2};
  int res = 0;
  do {
    st.clr();
    int c3 = n / 3;
    for (int i = 0; i < n;) {
      int pos = i;
      while (i < n && p[i].x == p[pos].x) {
        if (p[i].c == perm[2]) {
          --c3;
        } else {
          int tp = (p[i].c == perm[1]);
          st.update(1, 0, n - 1, p[i].x, tp);
        }
        ++i;
      }
      res = max(res, min(c3, st.get(1, 0, n - 1, 0, 0)));
    }
    for (int it = 0; it < 2; ++it) {
      st.clr();
      int c3 = n / 3;
      for (int i = 0; i < n;) {
        int pos = i;
        while (i < n && p[i].x == p[pos].x) {
          if (p[i].c == perm[2]) {
            --c3;
          } else {
            int tp = (p[i].c == perm[1]);
            st.update(1, 0, n - 1, p[i].y, tp);
          }
          ++i;
        }
        res = max(res, min(c3, st.get(1, 0, n - 1, 0, 0)));
      }
      for (int i = 0; i < n; ++i) {
        p[i].x = n - 1 - p[i].x;
      }
      reverse(p, p + n);
    }
  } while (next_permutation(perm.begin(), perm.end()));
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> p[i].x >> p[i].y >> p[i].c;
    --p[i].c;
    vx.push_back(p[i].x);
    vy.push_back(p[i].y);
  }
  sort(vx.begin(), vx.end());
  sort(vy.begin(), vy.end());
  vx.erase(unique(vx.begin(), vx.end()), vx.end());
  vy.erase(unique(vy.begin(), vy.end()), vy.end());
  for (int i = 0; i < n; ++i) {
    p[i].x = compress(vx, p[i].x);
    p[i].y = compress(vy, p[i].y);
  }
  int ans = 0;
  for (int it = 0; it < 2; ++it) {
    ans = max(ans, solve());
    for (int i = 0; i < n; ++i) {
      swap(p[i].x, p[i].y);
    }
  }
  cout << ans * 3 << "\n";
  return 0;
}

```