#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
int n, n1, n2, n3;
int x[N], y[N], c[N], co[N];
long long t[N];
int d1[N], d2[N];
inline int qry(int s) {
  int c1 = 0, c2 = 0, res = 0, i, pos = 0;
  for (i = (1 << 16); i; i >>= 1) {
    if (pos + i > n) continue;
    if (c1 + d1[pos + i] <= s - c2 - d2[pos + i]) {
      pos += i;
      c1 += d1[pos];
      c2 += d2[pos];
    }
  }
  res = min(c1, s - c2);
  if (pos < n) {
    c1 = 0;
    c2 = 0;
    for (i = pos + 1; i; i -= (i & (-i))) {
      c1 += d1[i];
      c2 += d2[i];
    }
    res = max(res, min(c1, s - c2));
  }
  return res;
}
pair<int, int> a1[N], a2[N], a3[N];
inline int solve1() {
  int i, j;
  n1 = n2 = n3 = 0;
  for (i = 1; i <= n; i++) {
    if (c[i] == 1) a1[++n1] = {x[i], y[i]};
    if (c[i] == 2) a2[++n2] = {x[i], y[i]};
    if (c[i] == 3) a3[++n3] = {x[i], y[i]};
  }
  sort(a1 + 1, a1 + n1 + 1);
  sort(a2 + 1, a2 + n2 + 1);
  sort(a3 + 1, a3 + n3 + 1);
  for (i = 1; i <= n; i++) d1[i] = d2[i] = 0;
  int p1 = 1, p2 = 1, res = 0;
  for (i = 1; i <= n3; i++) {
    while (p1 <= n1 && a1[p1].first < a3[i].first) {
      for (j = a1[p1].second; j <= n; j += (j & (-j))) d1[j]++;
      p1++;
    }
    while (p2 <= n2 && a2[p2].first < a3[i].first) {
      for (j = a2[p2].second; j <= n; j += (j & (-j))) d2[j]++;
      p2++;
    }
    res = max(res, min(n3 - i + 1, qry(p2 - 1)));
  }
  return res;
}
int b1[N], b2[N], b3[N];
inline int solve2() {
  int i, j;
  n1 = n2 = n3 = 0;
  for (i = 1; i <= n; i++) {
    if (c[i] == 1) b1[++n1] = x[i];
    if (c[i] == 2) b2[++n2] = x[i];
    if (c[i] == 3) b3[++n3] = x[i];
  }
  sort(b1 + 1, b1 + n1 + 1);
  sort(b2 + 1, b2 + n2 + 1);
  sort(b3 + 1, b3 + n3 + 1);
  for (i = 1; i <= n; i++) d1[i] = d2[i] = 0;
  int p1 = 1, p2 = 1, res = 0;
  for (i = 1; i <= n3; i++) {
    while (p1 <= n1 && b1[p1] < b3[i]) {
      for (j = b1[p1]; j <= n; j += (j & (-j))) d1[j]++;
      p1++;
    }
    while (p2 <= n2 && b2[p2] < b3[i]) {
      for (j = b2[p2]; j <= n; j += (j & (-j))) d2[j]++;
      p2++;
    }
    res = max(res, min(n3 - i + 1, qry(p2 - 1)));
  }
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int i, j;
  cin >> n;
  for (i = 1; i <= n; i++) cin >> x[i] >> y[i] >> co[i];
  for (i = 1; i <= n; i++) t[i] = x[i];
  sort(t + 1, t + n + 1);
  for (i = 1; i <= n; i++) x[i] = lower_bound(t + 1, t + n + 1, x[i]) - t;
  for (i = 1; i <= n; i++) t[i] = y[i];
  sort(t + 1, t + n + 1);
  for (i = 1; i <= n; i++) y[i] = lower_bound(t + 1, t + n + 1, y[i]) - t;
  int ans = 0;
  for (j = 1; j <= 4; j++) {
    int f[4] = {0, 1, 2, 3};
    do {
      for (i = 1; i <= n; i++) c[i] = f[co[i]];
      ans = max(ans, solve1());
      ans = max(ans, solve2());
    } while (next_permutation(f + 1, f + 4));
    for (i = 1; i <= n; i++) {
      swap(x[i], y[i]);
      y[i] = n - y[i] + 1;
    }
  }
  cout << ans * 3;
  return 0;
}

### 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;
const int N = 1e5 + 5;
int n, n1, n2, n3;
int x[N], y[N], c[N], co[N];
long long t[N];
int d1[N], d2[N];
inline int qry(int s) {
  int c1 = 0, c2 = 0, res = 0, i, pos = 0;
  for (i = (1 << 16); i; i >>= 1) {
    if (pos + i > n) continue;
    if (c1 + d1[pos + i] <= s - c2 - d2[pos + i]) {
      pos += i;
      c1 += d1[pos];
      c2 += d2[pos];
    }
  }
  res = min(c1, s - c2);
  if (pos < n) {
    c1 = 0;
    c2 = 0;
    for (i = pos + 1; i; i -= (i & (-i))) {
      c1 += d1[i];
      c2 += d2[i];
    }
    res = max(res, min(c1, s - c2));
  }
  return res;
}
pair<int, int> a1[N], a2[N], a3[N];
inline int solve1() {
  int i, j;
  n1 = n2 = n3 = 0;
  for (i = 1; i <= n; i++) {
    if (c[i] == 1) a1[++n1] = {x[i], y[i]};
    if (c[i] == 2) a2[++n2] = {x[i], y[i]};
    if (c[i] == 3) a3[++n3] = {x[i], y[i]};
  }
  sort(a1 + 1, a1 + n1 + 1);
  sort(a2 + 1, a2 + n2 + 1);
  sort(a3 + 1, a3 + n3 + 1);
  for (i = 1; i <= n; i++) d1[i] = d2[i] = 0;
  int p1 = 1, p2 = 1, res = 0;
  for (i = 1; i <= n3; i++) {
    while (p1 <= n1 && a1[p1].first < a3[i].first) {
      for (j = a1[p1].second; j <= n; j += (j & (-j))) d1[j]++;
      p1++;
    }
    while (p2 <= n2 && a2[p2].first < a3[i].first) {
      for (j = a2[p2].second; j <= n; j += (j & (-j))) d2[j]++;
      p2++;
    }
    res = max(res, min(n3 - i + 1, qry(p2 - 1)));
  }
  return res;
}
int b1[N], b2[N], b3[N];
inline int solve2() {
  int i, j;
  n1 = n2 = n3 = 0;
  for (i = 1; i <= n; i++) {
    if (c[i] == 1) b1[++n1] = x[i];
    if (c[i] == 2) b2[++n2] = x[i];
    if (c[i] == 3) b3[++n3] = x[i];
  }
  sort(b1 + 1, b1 + n1 + 1);
  sort(b2 + 1, b2 + n2 + 1);
  sort(b3 + 1, b3 + n3 + 1);
  for (i = 1; i <= n; i++) d1[i] = d2[i] = 0;
  int p1 = 1, p2 = 1, res = 0;
  for (i = 1; i <= n3; i++) {
    while (p1 <= n1 && b1[p1] < b3[i]) {
      for (j = b1[p1]; j <= n; j += (j & (-j))) d1[j]++;
      p1++;
    }
    while (p2 <= n2 && b2[p2] < b3[i]) {
      for (j = b2[p2]; j <= n; j += (j & (-j))) d2[j]++;
      p2++;
    }
    res = max(res, min(n3 - i + 1, qry(p2 - 1)));
  }
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int i, j;
  cin >> n;
  for (i = 1; i <= n; i++) cin >> x[i] >> y[i] >> co[i];
  for (i = 1; i <= n; i++) t[i] = x[i];
  sort(t + 1, t + n + 1);
  for (i = 1; i <= n; i++) x[i] = lower_bound(t + 1, t + n + 1, x[i]) - t;
  for (i = 1; i <= n; i++) t[i] = y[i];
  sort(t + 1, t + n + 1);
  for (i = 1; i <= n; i++) y[i] = lower_bound(t + 1, t + n + 1, y[i]) - t;
  int ans = 0;
  for (j = 1; j <= 4; j++) {
    int f[4] = {0, 1, 2, 3};
    do {
      for (i = 1; i <= n; i++) c[i] = f[co[i]];
      ans = max(ans, solve1());
      ans = max(ans, solve2());
    } while (next_permutation(f + 1, f + 4));
    for (i = 1; i <= n; i++) {
      swap(x[i], y[i]);
      y[i] = n - y[i] + 1;
    }
  }
  cout << ans * 3;
  return 0;
}

```