/*
	nota:
		Acest program nu rezolva problema, deoarece am citit prost:
		- in schimb face altceva(cred, sper): numarul minim de cutii ce trebuie selectate astfel incat
		sa putem baga restul cutiilor in ele

	* va urma si solutia la problema adevarata...
*/


#include <iostream>
#include <fstream>
#include <algorithm>
#include <stack>

using namespace std;

ifstream fin("cutii.in");
ofstream fout("cutii.out");

const int N = 3500;

struct pp{
	int x, y, z;
}v[N + 1];

bool cmp(pp a, pp b){
	if(a.x == b.x && a.y == b.y)
		return a.z > b.z;
	if(a.x = b.x)
		return a.y > b.y;
	return a.x > b.x;
}

struct AIB2d{
	int aib[N + 1][N + 1], n;

	AIB2d(){
		for(int i = 0; i <= N; i++)
			for(int j = 0; j <= N; j++)
				aib[i][j] = 0;
	}

	int lsb(int x){
		return x & -x;
	}

	void update(int iU, int jU, int val){
		for(int i = iU; i <= n; i += lsb(i))
			for(int j = jU; j <= n; j += lsb(j))
				aib[i][j] += val;
	}

	int query(int iQ, int jQ){
		int ans = 0;
		for(int i = iQ; i >= 1; i -= lsb(i))
			for(int j = jQ; j >= 1; j -= lsb(i))
				ans += aib[i][j];
		return ans;
	}

	void clear(){
		int val = query(1, 1);
		update(1, 1, -val);
	}
};

AIB2d myAib;

void solve(int n){
	myAib.clear();
	myAib.n = n;

	for(int i = 1; i <= n; i++)
		fin >> v[i].x >> v[i].y >> v[i].z;

	sort(v + 1, v + n + 1, cmp);

	int ans = 0;

	for(int i = 1; i <= n; i++){
		int val = myAib.query(n, n) + myAib.query(v[i].y, v[i].z)  
		        - myAib.query(v[i].y, n) - myAib.query(n, v[i].z)

		if(val > 0)
			myAib.update(v[i].y + 1, v[i].z + 1, -1);
		else
			ans++;
		myAib.update(v[i].y, v[i].z, 1);
	}

	fout << ans << '\n';
}

int main(){
	int n, T;
	fin >> n >> T;

	while(T--)
		solve(n);
	
	return 0;
}