#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 100;
int n, p[4], bx[N], wx, by[N], wy, sum[N][4];
vector<int> wh[N];
struct node {
  int x, y, c;
} sh[N];
bool cmp(node a, node b) { return a.x < b.x; }
inline int read() {
  int f = 1, x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = x * 10 + s - '0';
    s = getchar();
  }
  return x * f;
}
struct bit {
  int sh[N];
  void init() { memset(sh, 0, sizeof(sh)); }
  int lowbit(int x) { return x & (-x); }
  void change(int x, int v) {
    while (x <= n) sh[x] += v, x += lowbit(x);
  }
  int query(int x) {
    int ans = 0;
    while (x) ans += sh[x], x -= lowbit(x);
    return ans;
  }
  int ask(int l, int r) { return query(r) - query(l - 1); }
} T[4];
int g(int l, int r, int c) { return sum[r][c] - sum[l - 1][c]; }
void init() {
  memset(sum, 0, sizeof(sum));
  wx = wy = 0;
  for (int i = 1; i <= n; i++) bx[++wx] = sh[i].x;
  sort(bx + 1, bx + 1 + wx);
  wx = unique(bx + 1, bx + 1 + wx) - bx - 1;
  for (int i = 1; i <= n; i++)
    sh[i].x = lower_bound(bx + 1, bx + 1 + wx, sh[i].x) - bx;
  for (int i = 1; i <= n; i++) by[++wy] = sh[i].y;
  sort(by + 1, by + 1 + wy);
  wy = unique(by + 1, by + 1 + wy) - by - 1;
  for (int i = 1; i <= n; i++)
    sh[i].y = lower_bound(by + 1, by + 1 + wy, sh[i].y) - by;
  sort(sh + 1, sh + 1 + n, cmp);
  for (int i = 1; i <= n; i++) sum[sh[i].x][sh[i].c]++;
  for (int i = 1; i <= wx; i++)
    for (int j = 1; j <= 3; j++) sum[i][j] += sum[i - 1][j];
}
int solve() {
  for (int i = 1; i <= wx; i++) wh[i].clear();
  for (int i = 1; i <= n; i++) wh[sh[i].x].push_back(i);
  for (int i = 1; i <= 3; i++) p[i] = i;
  int ans = 1;
  do {
    for (int i = 2; i <= wx; i++) {
      int A = sum[i - 1][p[1]], l = i, r = wx - 1;
      while (l < r) {
        int mid = l + ((r - l + 1) >> 1);
        if (g(i, mid, p[2]) <= g(mid + 1, wx, p[3]))
          l = mid;
        else
          r = mid - 1;
      }
      ans = max(ans, min({A, g(i, l, p[2]), g(l + 1, wx, p[3])}));
      l++;
      ans = max(ans, min({A, g(i, l, p[2]), g(l + 1, wx, p[3])}));
    }
  } while (next_permutation(p + 1, p + 4));
  for (int i = 1; i <= 3; i++) p[i] = i;
  auto find = [](int A, int &ans) {
    int l = 1, r = wy - 1;
    while (l < r) {
      int mid = l + ((r - l + 1) >> 1);
      if (T[p[2]].ask(1, mid) <= T[p[3]].ask(mid + 1, wy))
        l = mid;
      else
        r = mid - 1;
    }
    ans = max(ans, min({A, T[p[2]].ask(1, l), T[p[3]].ask(l + 1, wy)}));
    l++;
    ans = max(ans, min({A, T[p[2]].ask(1, l), T[p[3]].ask(l + 1, wy)}));
  };
  do {
    T[p[2]].init();
    T[p[3]].init();
    for (int i = wx; i >= 1; i--) {
      int A = g(1, i, p[1]);
      find(A, ans);
      for (int j : wh[i])
        if (sh[j].c != p[1]) T[sh[j].c].change(sh[j].y, 1);
    }
  } while (next_permutation(p + 1, p + 4));
  for (int i = 1; i <= 3; i++) p[i] = i;
  do {
    T[p[2]].init();
    T[p[3]].init();
    for (int i = 1; i <= wx; i++) {
      int A = g(i, wx, p[1]);
      find(A, ans);
      for (int j : wh[i])
        if (sh[j].c != p[1]) T[sh[j].c].change(sh[j].y, 1);
    }
  } while (next_permutation(p + 1, p + 4));
  return ans;
}
signed main() {
  n = read();
  for (int i = 1; i <= n; i++) sh[i] = (node){read(), read(), read()};
  init();
  int ans = solve();
  for (int i = 1; i <= n; i++) swap(sh[i].x, sh[i].y);
  init();
  ans = max(ans, solve());
  printf("%d\n", ans * 3);
}

### Prompt

Develop a solution in Cpp 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 N = 1e5 + 100;
int n, p[4], bx[N], wx, by[N], wy, sum[N][4];
vector<int> wh[N];
struct node {
  int x, y, c;
} sh[N];
bool cmp(node a, node b) { return a.x < b.x; }
inline int read() {
  int f = 1, x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = x * 10 + s - '0';
    s = getchar();
  }
  return x * f;
}
struct bit {
  int sh[N];
  void init() { memset(sh, 0, sizeof(sh)); }
  int lowbit(int x) { return x & (-x); }
  void change(int x, int v) {
    while (x <= n) sh[x] += v, x += lowbit(x);
  }
  int query(int x) {
    int ans = 0;
    while (x) ans += sh[x], x -= lowbit(x);
    return ans;
  }
  int ask(int l, int r) { return query(r) - query(l - 1); }
} T[4];
int g(int l, int r, int c) { return sum[r][c] - sum[l - 1][c]; }
void init() {
  memset(sum, 0, sizeof(sum));
  wx = wy = 0;
  for (int i = 1; i <= n; i++) bx[++wx] = sh[i].x;
  sort(bx + 1, bx + 1 + wx);
  wx = unique(bx + 1, bx + 1 + wx) - bx - 1;
  for (int i = 1; i <= n; i++)
    sh[i].x = lower_bound(bx + 1, bx + 1 + wx, sh[i].x) - bx;
  for (int i = 1; i <= n; i++) by[++wy] = sh[i].y;
  sort(by + 1, by + 1 + wy);
  wy = unique(by + 1, by + 1 + wy) - by - 1;
  for (int i = 1; i <= n; i++)
    sh[i].y = lower_bound(by + 1, by + 1 + wy, sh[i].y) - by;
  sort(sh + 1, sh + 1 + n, cmp);
  for (int i = 1; i <= n; i++) sum[sh[i].x][sh[i].c]++;
  for (int i = 1; i <= wx; i++)
    for (int j = 1; j <= 3; j++) sum[i][j] += sum[i - 1][j];
}
int solve() {
  for (int i = 1; i <= wx; i++) wh[i].clear();
  for (int i = 1; i <= n; i++) wh[sh[i].x].push_back(i);
  for (int i = 1; i <= 3; i++) p[i] = i;
  int ans = 1;
  do {
    for (int i = 2; i <= wx; i++) {
      int A = sum[i - 1][p[1]], l = i, r = wx - 1;
      while (l < r) {
        int mid = l + ((r - l + 1) >> 1);
        if (g(i, mid, p[2]) <= g(mid + 1, wx, p[3]))
          l = mid;
        else
          r = mid - 1;
      }
      ans = max(ans, min({A, g(i, l, p[2]), g(l + 1, wx, p[3])}));
      l++;
      ans = max(ans, min({A, g(i, l, p[2]), g(l + 1, wx, p[3])}));
    }
  } while (next_permutation(p + 1, p + 4));
  for (int i = 1; i <= 3; i++) p[i] = i;
  auto find = [](int A, int &ans) {
    int l = 1, r = wy - 1;
    while (l < r) {
      int mid = l + ((r - l + 1) >> 1);
      if (T[p[2]].ask(1, mid) <= T[p[3]].ask(mid + 1, wy))
        l = mid;
      else
        r = mid - 1;
    }
    ans = max(ans, min({A, T[p[2]].ask(1, l), T[p[3]].ask(l + 1, wy)}));
    l++;
    ans = max(ans, min({A, T[p[2]].ask(1, l), T[p[3]].ask(l + 1, wy)}));
  };
  do {
    T[p[2]].init();
    T[p[3]].init();
    for (int i = wx; i >= 1; i--) {
      int A = g(1, i, p[1]);
      find(A, ans);
      for (int j : wh[i])
        if (sh[j].c != p[1]) T[sh[j].c].change(sh[j].y, 1);
    }
  } while (next_permutation(p + 1, p + 4));
  for (int i = 1; i <= 3; i++) p[i] = i;
  do {
    T[p[2]].init();
    T[p[3]].init();
    for (int i = 1; i <= wx; i++) {
      int A = g(i, wx, p[1]);
      find(A, ans);
      for (int j : wh[i])
        if (sh[j].c != p[1]) T[sh[j].c].change(sh[j].y, 1);
    }
  } while (next_permutation(p + 1, p + 4));
  return ans;
}
signed main() {
  n = read();
  for (int i = 1; i <= n; i++) sh[i] = (node){read(), read(), read()};
  init();
  int ans = solve();
  for (int i = 1; i <= n; i++) swap(sh[i].x, sh[i].y);
  init();
  ans = max(ans, solve());
  printf("%d\n", ans * 3);
}

```