#include <bits/stdc++.h>
using namespace std;
int n;
struct pnt {
  int x, y, id;
  pnt(int _x = 0, int _y = 0, int _id = 0) : x(_x), y(_y), id(_id) {}
};
bool xicmp(pnt a, pnt b) { return a.x < b.x; }
bool xdcmp(pnt a, pnt b) { return a.x > b.x; }
bool yicmp(pnt a, pnt b) { return a.y < b.y; }
bool ydcmp(pnt a, pnt b) { return a.y > b.y; }
pnt C[4][4][400000], P[1000000];
int cntc[4];
vector<int> V[4];
bool vis[4];
int xu, xd, yu, yd;
bool dfs(int cur, int x) {
  int XU = xu, XD = xd, YU = yu, YD = yd;
  V[cur].clear();
  for (int i = 1; i <= 3; i++) {
    if (!vis[i]) {
      if (cur == 3) {
        int cnt = 0;
        for (int j = 1; j <= n && cnt < x; j++)
          if (C[i][0][j].x > xd && C[i][0][j].x < xu && C[i][0][j].y > yd &&
              C[i][0][j].y < yu)
            cnt++, V[cur].push_back(C[i][0][j].id);
        if (cnt == x) return 1;
        V[cur].clear();
      } else {
        vis[i] = 1;
        for (int k = 0; k < 4; k++) {
          int cnt = 0;
          for (int j = 1; j <= n && cnt < x; j++)
            if (C[i][k][j].x > xd && C[i][k][j].x < xu && C[i][k][j].y > yd &&
                C[i][k][j].y < yu)
              cnt++, V[cur].push_back(C[i][k][j].id);
          if (cnt < x) {
            V[cur].clear();
            continue;
          }
          int u = V[cur][x - 1];
          if (k == 0)
            xd = max(xd, P[u].x);
          else if (k == 1)
            xu = min(xu, P[u].x);
          else if (k == 2)
            yd = max(yd, P[u].y);
          else
            yu = min(yu, P[u].y);
          if (dfs(cur + 1, x)) return 1;
          V[cur].clear();
          xd = XD, xu = XU, yd = YD, yu = YU;
        }
        vis[i] = 0;
      }
      V[cur].clear();
    }
  }
  return 0;
}
const int INF = 1e9 + 1;
bool check(int x) {
  for (int i = 1; i <= 3; i++) V[i].clear(), vis[i] = 0;
  xu = yu = INF, xd = yd = -INF;
  return dfs(1, x);
}
int main() {
  scanf("%d", &n);
  n /= 3;
  for (int i = 1, x = 0, y = 0, c = 0; i <= 3 * n; i++) {
    scanf("%d%d%d", &x, &y, &c);
    C[c][0][++cntc[c]] = (pnt(x, y, i));
    P[i].x = x, P[i].y = y, P[i].id = c;
  }
  for (int i = 1; i <= 3; i++)
    for (int j = 1; j <= 3; j++)
      for (int k = 1; k <= n; k++) C[i][j][k] = C[i][0][k];
  for (int i = 1; i <= 3; i++)
    sort(C[i][0] + 1, C[i][0] + n + 1, xicmp),
        sort(C[i][1] + 1, C[i][1] + n + 1, xdcmp),
        sort(C[i][2] + 1, C[i][2] + n + 1, yicmp),
        sort(C[i][3] + 1, C[i][3] + n + 1, ydcmp);
  int l = 0, r = n;
  while (l < r) {
    int mid = (l + r + 1) >> 1;
    if (check(mid))
      l = mid;
    else
      r = mid - 1;
  }
  printf("%d\n", 3 * l);
}

### 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;
int n;
struct pnt {
  int x, y, id;
  pnt(int _x = 0, int _y = 0, int _id = 0) : x(_x), y(_y), id(_id) {}
};
bool xicmp(pnt a, pnt b) { return a.x < b.x; }
bool xdcmp(pnt a, pnt b) { return a.x > b.x; }
bool yicmp(pnt a, pnt b) { return a.y < b.y; }
bool ydcmp(pnt a, pnt b) { return a.y > b.y; }
pnt C[4][4][400000], P[1000000];
int cntc[4];
vector<int> V[4];
bool vis[4];
int xu, xd, yu, yd;
bool dfs(int cur, int x) {
  int XU = xu, XD = xd, YU = yu, YD = yd;
  V[cur].clear();
  for (int i = 1; i <= 3; i++) {
    if (!vis[i]) {
      if (cur == 3) {
        int cnt = 0;
        for (int j = 1; j <= n && cnt < x; j++)
          if (C[i][0][j].x > xd && C[i][0][j].x < xu && C[i][0][j].y > yd &&
              C[i][0][j].y < yu)
            cnt++, V[cur].push_back(C[i][0][j].id);
        if (cnt == x) return 1;
        V[cur].clear();
      } else {
        vis[i] = 1;
        for (int k = 0; k < 4; k++) {
          int cnt = 0;
          for (int j = 1; j <= n && cnt < x; j++)
            if (C[i][k][j].x > xd && C[i][k][j].x < xu && C[i][k][j].y > yd &&
                C[i][k][j].y < yu)
              cnt++, V[cur].push_back(C[i][k][j].id);
          if (cnt < x) {
            V[cur].clear();
            continue;
          }
          int u = V[cur][x - 1];
          if (k == 0)
            xd = max(xd, P[u].x);
          else if (k == 1)
            xu = min(xu, P[u].x);
          else if (k == 2)
            yd = max(yd, P[u].y);
          else
            yu = min(yu, P[u].y);
          if (dfs(cur + 1, x)) return 1;
          V[cur].clear();
          xd = XD, xu = XU, yd = YD, yu = YU;
        }
        vis[i] = 0;
      }
      V[cur].clear();
    }
  }
  return 0;
}
const int INF = 1e9 + 1;
bool check(int x) {
  for (int i = 1; i <= 3; i++) V[i].clear(), vis[i] = 0;
  xu = yu = INF, xd = yd = -INF;
  return dfs(1, x);
}
int main() {
  scanf("%d", &n);
  n /= 3;
  for (int i = 1, x = 0, y = 0, c = 0; i <= 3 * n; i++) {
    scanf("%d%d%d", &x, &y, &c);
    C[c][0][++cntc[c]] = (pnt(x, y, i));
    P[i].x = x, P[i].y = y, P[i].id = c;
  }
  for (int i = 1; i <= 3; i++)
    for (int j = 1; j <= 3; j++)
      for (int k = 1; k <= n; k++) C[i][j][k] = C[i][0][k];
  for (int i = 1; i <= 3; i++)
    sort(C[i][0] + 1, C[i][0] + n + 1, xicmp),
        sort(C[i][1] + 1, C[i][1] + n + 1, xdcmp),
        sort(C[i][2] + 1, C[i][2] + n + 1, yicmp),
        sort(C[i][3] + 1, C[i][3] + n + 1, ydcmp);
  int l = 0, r = n;
  while (l < r) {
    int mid = (l + r + 1) >> 1;
    if (check(mid))
      l = mid;
    else
      r = mid - 1;
  }
  printf("%d\n", 3 * l);
}

```