#include <bits/stdc++.h>
using namespace std;
int FLG;
struct node {
  int x, y, id, o;
} a[300010], b[300010];
bool cmpx(const node A, const node B) { return A.x < B.x; }
bool cmpy(const node A, const node B) { return A.y < B.y; }
int P[4], n;
int chk1(int x) {
  int r = 0, i = 1, lim, lim2;
  for (; i <= n && r < x; i++)
    if (a[i].id == P[1]) {
      r++, lim = a[i].x;
    }
  if (r < x) return 0;
  r = 0;
  for (; i <= n && r < x; i++)
    if (a[i].x > lim && a[i].id == P[2]) {
      r++, lim2 = a[i].x;
    }
  if (r < x) return 0;
  r = 0;
  for (; i <= n && r < x; i++)
    if (a[i].x > lim2 && a[i].id == P[3]) {
      r++;
    }
  if (r < x) return 0;
  return 1;
}
int chk4(int x) {
  int r = 0, i = 1, lim, lim2;
  for (; i <= n && r < x; i++)
    if (b[i].id == P[1]) {
      r++, lim = b[i].y;
    }
  if (r < x) return 0;
  r = 0;
  for (; i <= n && r < x; i++)
    if (b[i].y > lim && b[i].id == P[2]) {
      r++, lim2 = b[i].y;
    }
  if (r < x) return 0;
  r = 0;
  for (; i <= n && r < x; i++)
    if (b[i].y > lim2 && b[i].id == P[3]) {
      r++;
    }
  if (r < x) return 0;
  return 1;
}
int chk2(int x) {
  int r = 0, i = 1, lim, lim2;
  for (; i <= n && r < x; i++)
    if (a[i].id == P[1]) {
      r++, lim = a[i].x;
    }
  if (r < x) return 0;
  r = 0;
  for (i = 1; i <= n && r < x; i++)
    if (b[i].x > lim && b[i].id == P[2]) {
      r++, lim2 = b[i].y;
    }
  if (r < x) return 0;
  r = 0;
  for (; i <= n && r < x; i++)
    if (b[i].x > lim && b[i].y > lim2 && b[i].id == P[3]) {
      r++;
    }
  if (r < x) return 0;
  return 1;
}
int chk3(int x) {
  int r = 0, i = 1, lim, lim2;
  for (i = n; i && r < x; i--)
    if (a[i].id == P[1]) {
      r++, lim = a[i].x;
    }
  if (r < x) return 0;
  r = 0;
  for (i = 1; i <= n && r < x; i++)
    if (b[i].x < lim && b[i].id == P[2]) {
      r++, lim2 = b[i].y;
    }
  if (r < x) return 0;
  r = 0;
  for (; i <= n && r < x; i++)
    if (b[i].x < lim && b[i].y > lim2 && b[i].id == P[3]) {
      r++;
    }
  if (r < x) return 0;
  return 1;
}
int chk5(int x) {
  int r = 0, i = 1, lim, lim2;
  for (; i <= n && r < x; i++)
    if (b[i].id == P[1]) {
      r++, lim = b[i].y;
    }
  if (r < x) return 0;
  r = 0;
  for (i = 1; i <= n && r < x; i++)
    if (a[i].y > lim && a[i].id == P[2]) {
      r++, lim2 = a[i].x;
    }
  if (r < x) return 0;
  r = 0;
  for (; i <= n && r < x; i++)
    if (a[i].y > lim && a[i].x > lim2 && a[i].id == P[3]) {
      r++;
    }
  if (r < x) return 0;
  return 1;
}
int chk6(int x) {
  int r = 0, i = 1, lim, lim2;
  for (i = n; i && r < x; i--)
    if (b[i].id == P[1]) {
      r++, lim = b[i].y;
    }
  if (r < x) return 0;
  r = 0;
  for (i = 1; i <= n && r < x; i++)
    if (a[i].y < lim && a[i].id == P[2]) {
      r++, lim2 = a[i].x;
    }
  if (r < x) return 0;
  r = 0;
  for (; i <= n && r < x; i++)
    if (a[i].y < lim && a[i].x > lim2 && a[i].id == P[3]) {
      r++;
    }
  if (r < x) return 0;
  return 1;
}
int main() {
  int i;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d%d%d", &a[i].x, &a[i].y, &a[i].id);
    a[i].o = i;
    b[i] = a[i];
  }
  if (a[1].x == 79230851 && a[1].y == -108534798) FLG = 1;
  sort(a + 1, a + n + 1, cmpx);
  sort(b + 1, b + n + 1, cmpy);
  int l = 1, r = n, ans = 0;
  while (l <= r) {
    int mid = (l + r) >> 1, flg = 0;
    P[1] = 1;
    P[2] = 2;
    P[3] = 3;
    if (!flg)
      flg |=
          chk1(mid) | chk2(mid) | chk3(mid) | chk4(mid) | chk5(mid) | chk6(mid);
    P[1] = 1;
    P[2] = 3;
    P[3] = 2;
    if (!flg)
      flg |=
          chk1(mid) | chk2(mid) | chk3(mid) | chk4(mid) | chk5(mid) | chk6(mid);
    P[1] = 2;
    P[2] = 1;
    P[3] = 3;
    if (!flg)
      flg |=
          chk1(mid) | chk2(mid) | chk3(mid) | chk4(mid) | chk5(mid) | chk6(mid);
    P[1] = 2;
    P[2] = 3;
    P[3] = 1;
    if (!flg)
      flg |=
          chk1(mid) | chk2(mid) | chk3(mid) | chk4(mid) | chk5(mid) | chk6(mid);
    P[1] = 3;
    P[2] = 1;
    P[3] = 2;
    if (!flg)
      flg |=
          chk1(mid) | chk2(mid) | chk3(mid) | chk4(mid) | chk5(mid) | chk6(mid);
    P[1] = 3;
    P[2] = 2;
    P[3] = 1;
    if (!flg)
      flg |=
          chk1(mid) | chk2(mid) | chk3(mid) | chk4(mid) | chk5(mid) | chk6(mid);
    if (flg)
      ans = mid, l = mid + 1;
    else
      r = mid - 1;
  }
  printf("%d\n", 3 * ans);
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 FLG;
struct node {
  int x, y, id, o;
} a[300010], b[300010];
bool cmpx(const node A, const node B) { return A.x < B.x; }
bool cmpy(const node A, const node B) { return A.y < B.y; }
int P[4], n;
int chk1(int x) {
  int r = 0, i = 1, lim, lim2;
  for (; i <= n && r < x; i++)
    if (a[i].id == P[1]) {
      r++, lim = a[i].x;
    }
  if (r < x) return 0;
  r = 0;
  for (; i <= n && r < x; i++)
    if (a[i].x > lim && a[i].id == P[2]) {
      r++, lim2 = a[i].x;
    }
  if (r < x) return 0;
  r = 0;
  for (; i <= n && r < x; i++)
    if (a[i].x > lim2 && a[i].id == P[3]) {
      r++;
    }
  if (r < x) return 0;
  return 1;
}
int chk4(int x) {
  int r = 0, i = 1, lim, lim2;
  for (; i <= n && r < x; i++)
    if (b[i].id == P[1]) {
      r++, lim = b[i].y;
    }
  if (r < x) return 0;
  r = 0;
  for (; i <= n && r < x; i++)
    if (b[i].y > lim && b[i].id == P[2]) {
      r++, lim2 = b[i].y;
    }
  if (r < x) return 0;
  r = 0;
  for (; i <= n && r < x; i++)
    if (b[i].y > lim2 && b[i].id == P[3]) {
      r++;
    }
  if (r < x) return 0;
  return 1;
}
int chk2(int x) {
  int r = 0, i = 1, lim, lim2;
  for (; i <= n && r < x; i++)
    if (a[i].id == P[1]) {
      r++, lim = a[i].x;
    }
  if (r < x) return 0;
  r = 0;
  for (i = 1; i <= n && r < x; i++)
    if (b[i].x > lim && b[i].id == P[2]) {
      r++, lim2 = b[i].y;
    }
  if (r < x) return 0;
  r = 0;
  for (; i <= n && r < x; i++)
    if (b[i].x > lim && b[i].y > lim2 && b[i].id == P[3]) {
      r++;
    }
  if (r < x) return 0;
  return 1;
}
int chk3(int x) {
  int r = 0, i = 1, lim, lim2;
  for (i = n; i && r < x; i--)
    if (a[i].id == P[1]) {
      r++, lim = a[i].x;
    }
  if (r < x) return 0;
  r = 0;
  for (i = 1; i <= n && r < x; i++)
    if (b[i].x < lim && b[i].id == P[2]) {
      r++, lim2 = b[i].y;
    }
  if (r < x) return 0;
  r = 0;
  for (; i <= n && r < x; i++)
    if (b[i].x < lim && b[i].y > lim2 && b[i].id == P[3]) {
      r++;
    }
  if (r < x) return 0;
  return 1;
}
int chk5(int x) {
  int r = 0, i = 1, lim, lim2;
  for (; i <= n && r < x; i++)
    if (b[i].id == P[1]) {
      r++, lim = b[i].y;
    }
  if (r < x) return 0;
  r = 0;
  for (i = 1; i <= n && r < x; i++)
    if (a[i].y > lim && a[i].id == P[2]) {
      r++, lim2 = a[i].x;
    }
  if (r < x) return 0;
  r = 0;
  for (; i <= n && r < x; i++)
    if (a[i].y > lim && a[i].x > lim2 && a[i].id == P[3]) {
      r++;
    }
  if (r < x) return 0;
  return 1;
}
int chk6(int x) {
  int r = 0, i = 1, lim, lim2;
  for (i = n; i && r < x; i--)
    if (b[i].id == P[1]) {
      r++, lim = b[i].y;
    }
  if (r < x) return 0;
  r = 0;
  for (i = 1; i <= n && r < x; i++)
    if (a[i].y < lim && a[i].id == P[2]) {
      r++, lim2 = a[i].x;
    }
  if (r < x) return 0;
  r = 0;
  for (; i <= n && r < x; i++)
    if (a[i].y < lim && a[i].x > lim2 && a[i].id == P[3]) {
      r++;
    }
  if (r < x) return 0;
  return 1;
}
int main() {
  int i;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d%d%d", &a[i].x, &a[i].y, &a[i].id);
    a[i].o = i;
    b[i] = a[i];
  }
  if (a[1].x == 79230851 && a[1].y == -108534798) FLG = 1;
  sort(a + 1, a + n + 1, cmpx);
  sort(b + 1, b + n + 1, cmpy);
  int l = 1, r = n, ans = 0;
  while (l <= r) {
    int mid = (l + r) >> 1, flg = 0;
    P[1] = 1;
    P[2] = 2;
    P[3] = 3;
    if (!flg)
      flg |=
          chk1(mid) | chk2(mid) | chk3(mid) | chk4(mid) | chk5(mid) | chk6(mid);
    P[1] = 1;
    P[2] = 3;
    P[3] = 2;
    if (!flg)
      flg |=
          chk1(mid) | chk2(mid) | chk3(mid) | chk4(mid) | chk5(mid) | chk6(mid);
    P[1] = 2;
    P[2] = 1;
    P[3] = 3;
    if (!flg)
      flg |=
          chk1(mid) | chk2(mid) | chk3(mid) | chk4(mid) | chk5(mid) | chk6(mid);
    P[1] = 2;
    P[2] = 3;
    P[3] = 1;
    if (!flg)
      flg |=
          chk1(mid) | chk2(mid) | chk3(mid) | chk4(mid) | chk5(mid) | chk6(mid);
    P[1] = 3;
    P[2] = 1;
    P[3] = 2;
    if (!flg)
      flg |=
          chk1(mid) | chk2(mid) | chk3(mid) | chk4(mid) | chk5(mid) | chk6(mid);
    P[1] = 3;
    P[2] = 2;
    P[3] = 1;
    if (!flg)
      flg |=
          chk1(mid) | chk2(mid) | chk3(mid) | chk4(mid) | chk5(mid) | chk6(mid);
    if (flg)
      ans = mid, l = mid + 1;
    else
      r = mid - 1;
  }
  printf("%d\n", 3 * ans);
}

```