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 <vector>
#include <list>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <fstream> 
#define _USE_MATH_DEFINES
#include <math.h>
#include <unordered_set>

using namespace std;


inline int toInt(string s) { int v; istringstream sin(s); sin >> v; return v; }
template<class T> inline string toString(T x) { ostringstream sout; sout << x; return sout.str(); }

template<class T> inline T sqr(T x) { return x*x; }

typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<long long int> vll;
typedef vector<string> vs;
typedef pair<int, int> pii;
typedef long long ll;
typedef unsigned long long ull;

//repetition
//------------------------------------------
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,n)  FOR(i,0,n)
#define P(p) cout<<(p)<<endl;
#define VEC_2D(a,b) vector<vector<int> >(a, vector<int>(b, 0))
#define ALL(a)  (a).begin(),(a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define pb push_back
#define mp make_pair
#define INF (1001000000)
#define SZ(a) int((a).size())
#define EACH(i,c) for(typeof((c).begin()) i=(c).begin(); i!=(c).end(); ++i)
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define SORT(c) sort((c).begin(),(c).end())
#define MOD 1000000007LL
#define FSP(a) cout << fixed << setprecision(a)


template<typename T> T gcd(T x, T y) {
	if (y == 0) return x;
	else return gcd(y, x%y);
}

template<typename T> T lcm(T a, T b) {
	return a / gcd(a, b) * b;
}
template<typename T> bool is_prime(T n) {
	for (int i = 2; i * i <= n; i++) {
		if (n % i == 0) return false;
	}
	return n != 1;
}
map<int, int> prime_factor(int n) {
	map<int, int> res;
	for (int i = 2; i * i <= n; i++) {
		while (n % i == 0) {
			++res[i];
			n /= i;
		}
	}
	if (n != 1) res[n] = 1;
	return res;
}
int extgcd(int a, int b, int& x, int& y) {//
	int d = a;
	if (b != 0) {
		d = extgcd(b, a%b, y, x);
		y -= (a / b)*x;
	}
	else {
		x = 1; y = 0;
	}
	return d;
}



ll mod_pow(ll x, ll n, ll mod) {
	if (n == 0) return 1;
	ll res = mod_pow(x * x % mod, n / 2, mod);
	if (n & 1) res = res * x % mod;
	return res;
}
vector<string> split(const string &str, char delim) {
	vector<string> res;
	size_t current = 0, found;
	while ((found = str.find_first_of(delim, current)) != string::npos) {
		res.push_back(string(str, current, found - current));
		current = found + 1;
	}
	res.push_back(string(str, current, str.size() - current));
	return res;
}

bool is_kadomatsu(int a, int b, int c) {
	if (a == b || a == c || b == c)return false;
	if (a > b && c > b) return true;
	if (a < b && c < b)return true;
	return false;
}


struct UF {
	int n;
	vi d;
	UF() {}
	UF(int n) :n(n), d(n, -1) {}
	int root(int v) {
		if (d[v] < 0) return v;
		return d[v] = root(d[v]);
	}
	bool same(int a, int b) {
		return root(a) == root(b);
	}
	bool unite(int x, int y) {
		x = root(x); y = root(y);
		if (x == y) return false;
		if (size(x) < size(y)) swap(x, y);
		d[x] += d[y];
		d[y] = x;
		return true;
	}
	int size(int v) { return -d[root(v)]; }
};



vector<int> divisor(int n) {
	if (n == 1) return{};
	vi res;
	for (int i = 1; i*i <= n; i++) {
		if (n%i == 0) {
			res.emplace_back(i);
			if (i != 1 && i != n / i)res.emplace_back(n / i);
		}
	}
	return res;
}

struct Bellmanford {
	int n;
	struct edge {
		int from, to, cost;
	};
	vector<edge> E;
	vi d;
	Bellmanford(int n) :n(n), d(n) {
E.resize(n);
	}
	void add_edge(int x, int y, int cost) {
		edge e;
		e.from = x; e.to = y; e.cost = cost;
		E.push_back(e);
	}
	void shortest_path(int s) {
		rep(i, n)d[i] = INF;
		d[s] = 0;
		while (true) {
			bool update = false;
			for (auto e : E) {
				if (d[e.from] != INF && d[e.to] > d[e.from] + e.cost) {
					d[e.to] = d[e.from] + e.cost;
					update = true;
				}
			}
			if (!update) break;
		}
	}
};
struct Dijkstra {
	int n;
	struct edge {
		int to; ll cost;
	};
	vector<vector<edge>> G;
	vll d;
	Dijkstra(int n) :n(n), d(n) {
		G.resize(n);
	}
	void add_edge(int x, int y, ll cost) {
		edge e;
		e.to = y; e.cost = cost;
		G[x].push_back(e);
	}
	void shortest_path(int s) {
		rep(i, n)d[i] = 100000000000000000;
		d[s] = 0;
		priority_queue<pair<ll, int>, vector<pair<ll, int>>, greater<pair<ll, int>>> que;
		que.push(make_pair(0, s));

		while (!que.empty()) {
			pii p = que.top(); que.pop();
			int v = p.second;
			if (d[v] < p.first) continue;
			for (auto e : G[v]) {
				if (d[e.to] > d[v] + e.cost) {
					d[e.to] = d[v] + e.cost;
					que.push(make_pair(d[e.to], e.to));
				}
			}
		}
	}
};

struct Segmenttree {
	int n;
	vector<pair<ll, int>> dat;
	Segmenttree() {}
	void init(ll n_) {
		n = 1;
		while (n < n_) n *= 2;
		dat.resize(2 * n - 1);
		rep(i, 2 * n - 1)dat[i] = pair<ll, int>(-INF, -INF);
	}
	void update(int idx, ll val) {
		idx += n - 1;
		dat[idx] = make_pair(val, -(idx - n + 1));
		while (idx > 0) {
			idx = (idx - 1) / 2;
			dat[idx] = max(dat[idx * 2 + 1], dat[idx * 2 + 2]);
		}
	}
	pair<ll, int> query(int a, int b) {
		return query_seg(a, b, 0, 0, n);
	}
	pair<ll, int> query_seg(int a, int b, int k, int l, int r) {
		if (r <= a || b <= l) return pair<ll, int>(-INF, -INF);
		if (a <= l && r <= b)return dat[k];
		else {
			return max(query_seg(a, b, k * 2 + 1, l, (l + r) / 2), query_seg(a, b, k * 2 + 2, (l + r) / 2, r));
		}
	}
};
//------------------------------------------------------------


int main() {

	int n, c;
	cin >> n >> c;
	vi s(n), t(n), cc(n);
	rep(i, n)cin >> s[i] >> t[i] >> cc[i];

	int tv[30][100005];
	memset(tv, 0, sizeof(tv));
	rep(i, n) {
		tv[cc[i]-1][s[i]] += 1;
		tv[cc[i]-1][t[i] + 1] -= 1;
	}
	rep(i, 30) {
		for (int j = 1; j < 100005; j++) {
			tv[i][j] += tv[i][j - 1];
		}
	}
	int ans = 0;
	rep(i, 100005) {
		int count = 0;
		rep(j, 30) {
			if (tv[j][i] > 0)count++;
		}
		ans = max(ans, count);
	}
	P(ans);
	return 0;
}