#include <bits/stdc++.h>
using namespace std;
template <typename T>
bool chkMax(T &x, T y) {
  return (y > x) ? x = y, 1 : 0;
}
template <typename T>
bool chkMin(T &x, T y) {
  return (y < x) ? x = y, 1 : 0;
}
template <typename T>
void inline read(T &x) {
  int f = 1;
  x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s <= '9' && s >= '0') x = x * 10 + (s ^ 48), s = getchar();
  x *= f;
}
const int N = 3e5 + 5;
int n, t[3], ans, p[3], m, ret[N];
struct E {
  int x, y, id;
} e[3][N], g[3][N], px[3][N], py[3][N];
bool inline cmpX(E a, E b) { return a.x < b.x; }
bool inline cmpY(E a, E b) { return a.y < b.y; }
void inline tr(int k) {
  for (int i = 1; i <= k; i++) ret[i] = g[0][i].id;
  int p = g[0][k].x + 1, cnt = 0, U, D;
  for (int i = 1; i <= m; i++) {
    if (g[1][i].x >= p) {
      cnt++;
      ret[cnt + k] = g[1][i].id;
      if (cnt == k) break;
    }
  }
  cnt = 0;
  for (int i = m; i; i--) {
    if (g[2][i].x >= p) {
      cnt++;
      D = g[2][i].y;
      ret[cnt + k * 2] = g[2][i].id;
      if (cnt == k) break;
    }
  }
}
bool inline chk(int k) {
  int p = g[0][k].x + 1;
  int U, D, cnt = 0;
  for (int i = 1; i <= m; i++) {
    if (g[1][i].x >= p) {
      cnt++;
      U = g[1][i].y;
      if (cnt == k) break;
    }
  }
  if (cnt != k) return 0;
  cnt = 0;
  for (int i = m; i; i--) {
    if (g[2][i].x >= p) {
      cnt++;
      D = g[2][i].y;
      if (cnt == k) break;
    }
  }
  if (cnt != k) return 0;
  return U < D;
}
void inline solve() {
  for (int i = 1; i <= m; i++) {
    g[0][i] = px[p[0]][i];
    g[1][i] = py[p[1]][i];
    g[2][i] = py[p[2]][i];
  }
  int l = ans, r = m;
  while (l < r) {
    int mid = (l + r + 1) >> 1;
    if (chk(mid))
      l = mid;
    else
      r = mid - 1;
  }
  if (chk(r) && chkMax(ans, r)) tr(r);
}
void inline tr2(int k) {
  int p = g[0][k].x + 1, q = g[2][m - k + 1].x - 1;
  for (int i = 1; i <= k; i++)
    ret[i] = g[0][i].id, ret[i + k] = g[2][m - k + i].id;
  int U, D, cnt = 0;
  for (int i = 1; i <= m; i++) {
    if (g[1][i].x >= p && g[1][i].x <= q) {
      cnt++;
      ret[cnt + k * 2] = g[1][i].id;
      if (cnt == k) break;
    }
  }
}
bool inline ck2(int k) {
  int p = g[0][k].x + 1, q = g[2][m - k + 1].x - 1;
  int U, D, cnt = 0;
  for (int i = 1; i <= m; i++) {
    if (g[1][i].x >= p && g[1][i].x <= q) {
      cnt++;
      if (cnt == k) break;
    }
  }
  if (cnt != k) return 0;
  return 1;
}
void inline solve2() {
  for (int i = 1; i <= m; i++) {
    g[0][i] = px[p[0]][i];
    g[1][i] = px[p[1]][i];
    g[2][i] = px[p[2]][i];
  }
  int l = ans, r = m;
  while (l < r) {
    int mid = (l + r + 1) >> 1;
    if (ck2(mid))
      l = mid;
    else
      r = mid - 1;
  }
  if (ck2(r) && chkMax(ans, r)) tr2(r);
}
void inline work() {
  for (int i = 0; i < 3; i++) {
    sort(px[i] + 1, px[i] + 1 + m, cmpX);
    sort(py[i] + 1, py[i] + 1 + m, cmpY);
  }
  p[0] = 0, p[1] = 1, p[2] = 2;
  do {
    solve();
  } while (next_permutation(p, p + 3));
}
void inline WK() {
  for (int i = 0; i < 3; i++) {
    sort(px[i] + 1, px[i] + 1 + m, cmpX);
    sort(py[i] + 1, py[i] + 1 + m, cmpY);
  }
  p[0] = 0, p[1] = 1, p[2] = 2;
  do {
    solve2();
  } while (next_permutation(p, p + 3));
}
void inline dododo() {
  for (int i = 0; i < 3; i++) {
    sort(px[i] + 1, px[i] + 1 + m, cmpX);
    sort(py[i] + 1, py[i] + 1 + m, cmpY);
  }
  WK();
  work();
  for (int i = 1; i <= m; i++)
    for (int j = 0; j < 3; j++) px[j][i].x *= -1, py[j][i].x *= -1;
  work();
  for (int i = 1; i <= m; i++)
    for (int j = 0; j < 3; j++) px[j][i].y *= -1, py[j][i].y *= -1;
  work();
  for (int i = 1; i <= m; i++)
    for (int j = 0; j < 3; j++) px[j][i].x *= -1, py[j][i].x *= -1;
  work();
}
int main() {
  read(n);
  m = n / 3;
  for (int i = 1; i <= n; i++) {
    int x, y, c;
    read(x), read(y), read(c);
    c--;
    e[c][++t[c]] = (E){x, y, i};
  }
  for (int i = 1; i <= m; i++)
    for (int j = 0; j < 3; j++) px[j][i] = py[j][i] = e[j][i];
  dododo();
  for (int i = 1; i <= m; i++)
    for (int j = 0; j < 3; j++) swap(e[j][i].x, e[j][i].y);
  for (int i = 1; i <= m; i++)
    for (int j = 0; j < 3; j++) px[j][i] = py[j][i] = e[j][i];
  dododo();
  printf("%d\n", ans * 3);
  return 0;
}

### Prompt

Your task is to create 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;
template <typename T>
bool chkMax(T &x, T y) {
  return (y > x) ? x = y, 1 : 0;
}
template <typename T>
bool chkMin(T &x, T y) {
  return (y < x) ? x = y, 1 : 0;
}
template <typename T>
void inline read(T &x) {
  int f = 1;
  x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s <= '9' && s >= '0') x = x * 10 + (s ^ 48), s = getchar();
  x *= f;
}
const int N = 3e5 + 5;
int n, t[3], ans, p[3], m, ret[N];
struct E {
  int x, y, id;
} e[3][N], g[3][N], px[3][N], py[3][N];
bool inline cmpX(E a, E b) { return a.x < b.x; }
bool inline cmpY(E a, E b) { return a.y < b.y; }
void inline tr(int k) {
  for (int i = 1; i <= k; i++) ret[i] = g[0][i].id;
  int p = g[0][k].x + 1, cnt = 0, U, D;
  for (int i = 1; i <= m; i++) {
    if (g[1][i].x >= p) {
      cnt++;
      ret[cnt + k] = g[1][i].id;
      if (cnt == k) break;
    }
  }
  cnt = 0;
  for (int i = m; i; i--) {
    if (g[2][i].x >= p) {
      cnt++;
      D = g[2][i].y;
      ret[cnt + k * 2] = g[2][i].id;
      if (cnt == k) break;
    }
  }
}
bool inline chk(int k) {
  int p = g[0][k].x + 1;
  int U, D, cnt = 0;
  for (int i = 1; i <= m; i++) {
    if (g[1][i].x >= p) {
      cnt++;
      U = g[1][i].y;
      if (cnt == k) break;
    }
  }
  if (cnt != k) return 0;
  cnt = 0;
  for (int i = m; i; i--) {
    if (g[2][i].x >= p) {
      cnt++;
      D = g[2][i].y;
      if (cnt == k) break;
    }
  }
  if (cnt != k) return 0;
  return U < D;
}
void inline solve() {
  for (int i = 1; i <= m; i++) {
    g[0][i] = px[p[0]][i];
    g[1][i] = py[p[1]][i];
    g[2][i] = py[p[2]][i];
  }
  int l = ans, r = m;
  while (l < r) {
    int mid = (l + r + 1) >> 1;
    if (chk(mid))
      l = mid;
    else
      r = mid - 1;
  }
  if (chk(r) && chkMax(ans, r)) tr(r);
}
void inline tr2(int k) {
  int p = g[0][k].x + 1, q = g[2][m - k + 1].x - 1;
  for (int i = 1; i <= k; i++)
    ret[i] = g[0][i].id, ret[i + k] = g[2][m - k + i].id;
  int U, D, cnt = 0;
  for (int i = 1; i <= m; i++) {
    if (g[1][i].x >= p && g[1][i].x <= q) {
      cnt++;
      ret[cnt + k * 2] = g[1][i].id;
      if (cnt == k) break;
    }
  }
}
bool inline ck2(int k) {
  int p = g[0][k].x + 1, q = g[2][m - k + 1].x - 1;
  int U, D, cnt = 0;
  for (int i = 1; i <= m; i++) {
    if (g[1][i].x >= p && g[1][i].x <= q) {
      cnt++;
      if (cnt == k) break;
    }
  }
  if (cnt != k) return 0;
  return 1;
}
void inline solve2() {
  for (int i = 1; i <= m; i++) {
    g[0][i] = px[p[0]][i];
    g[1][i] = px[p[1]][i];
    g[2][i] = px[p[2]][i];
  }
  int l = ans, r = m;
  while (l < r) {
    int mid = (l + r + 1) >> 1;
    if (ck2(mid))
      l = mid;
    else
      r = mid - 1;
  }
  if (ck2(r) && chkMax(ans, r)) tr2(r);
}
void inline work() {
  for (int i = 0; i < 3; i++) {
    sort(px[i] + 1, px[i] + 1 + m, cmpX);
    sort(py[i] + 1, py[i] + 1 + m, cmpY);
  }
  p[0] = 0, p[1] = 1, p[2] = 2;
  do {
    solve();
  } while (next_permutation(p, p + 3));
}
void inline WK() {
  for (int i = 0; i < 3; i++) {
    sort(px[i] + 1, px[i] + 1 + m, cmpX);
    sort(py[i] + 1, py[i] + 1 + m, cmpY);
  }
  p[0] = 0, p[1] = 1, p[2] = 2;
  do {
    solve2();
  } while (next_permutation(p, p + 3));
}
void inline dododo() {
  for (int i = 0; i < 3; i++) {
    sort(px[i] + 1, px[i] + 1 + m, cmpX);
    sort(py[i] + 1, py[i] + 1 + m, cmpY);
  }
  WK();
  work();
  for (int i = 1; i <= m; i++)
    for (int j = 0; j < 3; j++) px[j][i].x *= -1, py[j][i].x *= -1;
  work();
  for (int i = 1; i <= m; i++)
    for (int j = 0; j < 3; j++) px[j][i].y *= -1, py[j][i].y *= -1;
  work();
  for (int i = 1; i <= m; i++)
    for (int j = 0; j < 3; j++) px[j][i].x *= -1, py[j][i].x *= -1;
  work();
}
int main() {
  read(n);
  m = n / 3;
  for (int i = 1; i <= n; i++) {
    int x, y, c;
    read(x), read(y), read(c);
    c--;
    e[c][++t[c]] = (E){x, y, i};
  }
  for (int i = 1; i <= m; i++)
    for (int j = 0; j < 3; j++) px[j][i] = py[j][i] = e[j][i];
  dododo();
  for (int i = 1; i <= m; i++)
    for (int j = 0; j < 3; j++) swap(e[j][i].x, e[j][i].y);
  for (int i = 1; i <= m; i++)
    for (int j = 0; j < 3; j++) px[j][i] = py[j][i] = e[j][i];
  dododo();
  printf("%d\n", ans * 3);
  return 0;
}

```