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: Accepted)

#include <stdio.h>
#include <iostream>
#include <cstdio>
#include <algorithm> 
#include <stdio.h>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <cfloat>
#include <map>
#include <list>
#include <utility>
#include <set>
#include <memory>
#include <string>
#include <vector>
#include <algorithm>
#include <functional>
#include <sstream>
#include <complex>
#include <stack>
#include <queue>
#include <numeric>
#include <bitset>
#include <stdio.h>
#include <string.h>
#include <assert.h>
#include <random>

using namespace std;

static const double EPS = 1e-9;
template<class T> bool INRANGE(T x, T a, T b) { return a <= x && x <= b; }
template<class T> void amin(T &a, T v) { if (a > v) a = v; }
template<class T> void amax(T &a, T v) { if (a < v) a = v; }
int ROUND(double x) { return (int)(x + 0.5); }
bool ISINT(double x) { return fabs(ROUND(x) - x) <= EPS; }
bool ISEQUAL(double x, double y) { return fabs(x - y) <= EPS * max(1.0, max(fabs(x), fabs(y))); }
double SQSUM(double x, double y) { return x * x + y * y; }
template<class T>
vector<T> make_vector(int n, T t) {
	return vector<T>(n, t);
}
 
template<class ...Ts>
auto make_vector(int n, Ts ... ts) {
	return vector<decltype(make_vector(ts...))>(n, make_vector(ts...));
}

#define PI  (acos(-1))
#define ARRAY_NUM(a) (sizeof(a)/sizeof(a[0])) 
#define NG (-1)
#define BIG ((int)1e9+10)
#define BIGLL ((ll)4e18)
#define SZ(a) ((int)(a).size())
#define SQ(a) ((a)*(a))
#define REP(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define ALL(x) (x).begin(),(x).end()
#define YES(n) cout << ((n) ? "YES" : "NO"  ) << endl
#define Yes(n) cout << ((n) ? "Yes" : "No"  ) << endl
#define POSSIBLE(n) cout << ((n) ? "POSSIBLE" : "IMPOSSIBLE"  ) << endl
#define Possible(n) cout << ((n) ? "Possible" : "Impossible"  ) << endl

#define BitSet(arg,posn) ((arg) |= (1LL << (posn)))
#define BitClr(arg,posn) ((arg) &= ~(1LL << (posn)))
#define BitFlp(arg,posn) ((arg) ^= (1LL << (posn)))
#define IsBit(arg,posn) static_cast<bool>((arg) & (1LL << (posn)))

// #define DEBUG(x) cerr<<#x<<": "<<(x)<<endl
// #define DEBUG_VEC(v) cerr<<#v<<": ";REP(__i,(v).size())cerr<<((v)[__i])<<", ";cerr<<endl

typedef unsigned long long ull;
typedef long long ll;

struct in {
	int n;
	in() {}
	in(int n_) : n(n_) {};
	template <class T> operator T() {
		T ret;
		cin >> ret;
		return ret;
	}
	template <class T> operator vector<T>() {
		vector<T> ret(n);
		for (int i = 0; i < n; i++) cin >> ret[i];
		return ret;
	}
};

// modint: mod 計算を int を扱うように扱える構造体
template<int MOD>
struct ModInt {
	long long val;
	constexpr ModInt(long long v = 0) noexcept : val(v % MOD) {
		if (val < 0) v += MOD;
	}
	constexpr int getmod() { return MOD; }
	constexpr ModInt operator - () const noexcept {
		return val ? MOD - val : 0;
	}
	constexpr ModInt operator + (const ModInt& r) const noexcept { return ModInt(*this) += r; }
	constexpr ModInt operator - (const ModInt& r) const noexcept { return ModInt(*this) -= r; }
	constexpr ModInt operator * (const ModInt& r) const noexcept { return ModInt(*this) *= r; }
	constexpr ModInt operator / (const ModInt& r) const noexcept { return ModInt(*this) /= r; }
	constexpr ModInt& operator += (const ModInt& r) noexcept {
		val += r.val;
		if (val >= MOD) val -= MOD;
		return *this;
	}
	constexpr ModInt& operator -= (const ModInt& r) noexcept {
		val -= r.val;
		if (val < 0) val += MOD;
		return *this;
	}
	constexpr ModInt& operator *= (const ModInt& r) noexcept {
		val = val * r.val % MOD;
		return *this;
	}
	constexpr ModInt& operator /= (const ModInt& r) noexcept {
		long long a = r.val, b = MOD, u = 1, v = 0;
		while (b) {
			long long t = a / b;
			a -= t * b; swap(a, b);
			u -= t * v; swap(u, v);
		}
		val = val * u % MOD;
		if (val < 0) val += MOD;
		return *this;
	}
	constexpr bool operator == (const ModInt& r) const noexcept {
		return this->val == r.val;
	}
	constexpr bool operator != (const ModInt& r) const noexcept {
		return this->val != r.val;
	}

	friend ostream& operator << (ostream &os, const ModInt<MOD>& x) noexcept {
		return os << x.val;
	}
	friend istream& operator >> (istream &is, ModInt<MOD>& x) noexcept {
		return is >> x.val;
	}
	friend constexpr ModInt<MOD> modpow(const ModInt<MOD> &a, long long n) noexcept {
		if (n == 0) return 1;
		auto t = modpow(a, n / 2);
		t = t * t;
		if (n & 1) t = t * a;
		return t;
	}
};

// 二項係数ライブラリ
template<class T> struct BiCoef
{
	constexpr BiCoef() {}
	constexpr BiCoef(int n) noexcept : mFact(n, 1), mInv(n, 1), mFInv(n, 1)
	{
		Init(n);
	}

	constexpr void Init(int n) noexcept
	{
		mFact.assign(n, 1), mInv.assign(n, 1), mFInv.assign(n, 1);
		int factMod = mFact[0].getmod();
		for (int i = 2; i < n; i++) {
			mFact[i] = mFact[i - 1] * i;
			mInv[i] = -mInv[factMod%i] * (factMod / i);
			mFInv[i] = mFInv[i - 1] * mInv[i];
		}
	}

	constexpr T Fact(int n) const noexcept {
		if (n < 0) return 0;
		return mFact[n];
	}

	constexpr T Comb(int n, int k) const noexcept {
		if (n < k || n < 0 || k < 0) return 0;
		return mFact[n] * mFInv[k] * mFInv[n - k];
	}

	constexpr T Perm(int n, int k) const noexcept {
		if (n < k || n < 0 || k < 0) return 0;
		return mFact[n] * mFInv[k] * mFInv[n - k];
	}

	constexpr T Homo(int n, int r) const {
		if (n < 0 || r < 0) return 0;
		return r == 0 ? 1 : Comb(n + r - 1, r);
	}

	constexpr T inv(int n) const noexcept {
		if (n < 0) return 0;
		return mInv[n];
	}

	constexpr T finv(int n) const noexcept {
		if (n < 0) return 0;
		return mFInv[n];
	}

	vector<T> mFact;
	vector<T> mInv;
	vector<T> mFInv;
};

const int MOD = 1000000007;
using mint = ModInt<MOD>;
// BiCoef<mint> bc(1234567); // 1M * 3 * 8B(ll) = 24MB 

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(nullptr);

	ll N, C;
	cin >> N >> C;

	vector <int> s, t, c;

	REP(i,N)
	{
		int tmp;
		cin >> tmp;
		tmp *= 2;
		s.push_back(tmp);
		cin >> tmp;
		tmp *= 2;
		t.push_back(tmp);
		cin >> tmp;
		tmp--;
		c.push_back(tmp);
	}

	ll sum[200005] = {};
	REP(nowC, C)
	{
		ll line[200005] = {};
		REP(i, N)
		{
			if (nowC == c[i])
			{
				line[s[i] - 1]++;
				line[t[i]]--;
			}
		}

		REP(m, 200001)
		{
			line[m + 1] += line[m];
		}

		REP(m, 200001)
		{
			if (line[m])
			{
				sum[m]++;
			}
		}
	}

	ll ret = *max_element(sum, sum + 200005);
	cout << ret << endl;





	return 0;
}