Problem:
Score : 400 points

Problem Statement

Joisino is planning to record N TV programs with recorders.

The TV can receive C channels numbered 1 through C.

The i-th program that she wants to record will be broadcast from time s_i to time t_i (including time s_i but not t_i) on Channel c_i.

Here, there will never be more than one program that are broadcast on the same channel at the same time.

When the recorder is recording a channel from time S to time T (including time S but not T), it cannot record other channels from time S-0.5 to time T (including time S-0.5 but not T).

Find the minimum number of recorders required to record the channels so that all the N programs are completely recorded.

Constraints

1≤N≤10^5

1≤C≤30

1≤s_i<t_i≤10^5

1≤c_i≤C

If c_i=c_j and i≠j, either t_i≤s_j or s_i≥t_j.

All input values are integers.

Input

Input is given from Standard Input in the following format:

N C
s_1 t_1 c_1
:
s_N t_N c_N

Output

When the minimum required number of recorders is x, print the value of x.

Sample Input 1

3 2
1 7 2
7 8 1
8 12 1

Sample Output 1

2

Two recorders can record all the programs, for example, as follows:

With the first recorder, record Channel 2 from time 1 to time 7. The first program will be recorded. Note that this recorder will be unable to record other channels from time 0.5 to time 7.

With the second recorder, record Channel 1 from time 7 to time 12. The second and third programs will be recorded. Note that this recorder will be unable to record other channels from time 6.5 to time 12.

Sample Input 2

3 4
1 3 2
3 4 4
1 4 3

Sample Output 2

3

There may be a channel where there is no program to record.

Sample Input 3

9 4
56 60 4
33 37 2
89 90 3
32 43 1
67 68 3
49 51 3
31 32 3
70 71 1
11 12 3

Sample Output 3

2

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

//////////////////////////////   Begin Macros

#define all(x) (x).begin(), (x).end()
#define rep(i, N) for (int i = 0; i < (int)(N); i++)
#define reprev(i, N) for (int i = (int)(N)-1; i >= 0; i--)
#define rep1(i, N) for (int i = 1; i <= (int)(N); i++)
#define rep1rev(i, N) for (int i = (int)(N); i >= 0; i--)
#define forbe(i, b, e) for (int i = (b); i < (e); i++)
#define forberev(i, b, e) for (int i = (e)-1; i >= (b); i--)
#define forfl(i, f, l) for (int i = (f); i <= (l); i++)
#define forflrev(i, f, l) for (int i = (l); i >= (f); i--)
using ll = long long int;
using pii = pair<int, int>;
using pil = pair<int, ll>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;

template <typename T>
bool chmax(T &m, const T q)
{
  if (m < q)
  {
    m = q;
    return true;
  }
  else
    return false;
}
template <typename T>
bool chmin(T &m, const T q)
{
  if (m > q)
  {
    m = q;
    return true;
  }
  else
    return false;
}
template <typename T1, typename T2>
pair<T1, T2> operator+(const pair<T1, T2> &l, const pair<T1, T2> &r) { return {l.first + r.first, l.second + r.second}; }
template <typename T1, typename T2>
pair<T1, T2> operator-(const pair<T1, T2> &l, const pair<T1, T2> &r) { return {l.first - r.first, l.second - r.second}; }
template <typename T>
pair<T, T> operator*(const pair<T, T> &l, const T &r) { return {l.first * r, l.second * r}; }
template <typename T>
pair<T, T> operator/(const pair<T, T> &l, const T &r) { return {l.first / r, l.second / r}; }
template <typename T>
istream &operator>>(istream &is, vector<T> &vec)
{
  for (auto &v : vec)
    is >> v;
  return is;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &vec)
{
  os << "[";
  for (auto v : vec)
    os << v << ",";
  os << "]";
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const deque<T> &vec)
{
  os << "deq[";
  for (auto v : vec)
    os << v << ",";
  os << "]";
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &vec)
{
  os << "{";
  for (auto v : vec)
    os << v << ",";
  os << "}";
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const unordered_set<T> &vec)
{
  os << "{";
  for (auto v : vec)
    os << v << ",";
  os << "}";
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const multiset<T> &vec)
{
  os << "{";
  for (auto v : vec)
    os << v << ",";
  os << "}";
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const unordered_multiset<T> &vec)
{
  os << "{";
  for (auto v : vec)
    os << v << ",";
  os << "}";
  return os;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &pa)
{
  os << "(" << pa.first << "," << pa.second << ")";
  return os;
}
template <typename TK, typename TV>
ostream &operator<<(ostream &os, const map<TK, TV> &mp)
{
  os << "{";
  for (auto v : mp)
    os << v.first << "=>" << v.second << ",";
  os << "}";
  return os;
}
template <typename TK, typename TV>
ostream &operator<<(ostream &os, const unordered_map<TK, TV> &mp)
{
  os << "{";
  for (auto v : mp)
    os << v.first << "=>" << v.second << ",";
  os << "}";
  return os;
}

template <typename T>
void reset(vector<T> &v, const T reset_to)
{
  for (auto &x : v)
    x = reset_to;
}
inline int popcount(const unsigned int x) { return __builtin_popcount(x); }
#define dbg(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ") " << __FILE__ << endl;

ll nC2(ll n)
{
  return n * (n - 1) / 2;
}

const int intinf = numeric_limits<int>::max();
const ll llinf = numeric_limits<ll>::max();
const pii udlr[4] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};

//////////////////////////////   End Macros

void solve()
{
  int N, C;
  cin >> N >> C;
  vector<int> s(N), t(N), c(N + 1);
  rep(i, N) cin >> s[i] >> t[i] >> c[i];
  rep(i, N) c[i]--;

  vector<vector<pii>> st(C);
  rep(i, N) st[c[i]].push_back({s[i], t[i]});

  vector<pii> brks;
  rep(j, C)
  {
    int S, T;
    S = 0;
    T = 0;
    for (auto st_ : st[j])
    {
      int s_ = st_.first;
      int t_ = st_.second;
      if (s_ == T)
      {
        T = t_;
        continue;
      }

      if (T != 0)
        brks.push_back({T, 1});
      brks.push_back({S = s_, 0});
      T = t_;
    }
    if (T != 0)
      brks.push_back({T, 1});
  }
  sort(all(brks));
  dbg(brks);

  int overlap = 0;
  int ans = 0;
  for (auto brk : brks)
  {
    if (brk.second == 0)
    {
      overlap++;
      chmax(ans, overlap);
    }
    else
      overlap--;
  }
  cout << ans << endl;
}

int main()
{
  cout << fixed << setprecision(15);
  solve();
  return 0;
}