平面上有n个点，每个点有k种颜色中的一个。
你可以选择一条水平的线段获得在其上方或其下方的所有点，如图所示：

 

 

请求出你最多能够得到多少点，使得获得的点并不包含所有的颜色。

包含多组测试数据，第一行输入一个数T表示测试数据组数。
接下来T组测试数据，对于每组测试数据，第一行输入两个数n，k，分别表示点的个数和颜色数。
接下来n行每行描述一个点，前两个数z，y(lxl，lyl≤2^32-1)描述点的位置，最后一个数z(1≤z≤K）描述点的颜色。

对于每组数据输出一行，每行一个数ans，表示答案。

1
10 3
1 2 3
2 1 1
2 4 2
3 5 3
4 4 2
5 1 2
6 3 1
6 7 1
7 2 3
9 4 2

5

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <map>
#include <set>
using namespace std;

const int BufferSize = 1 << 16;
char buffer[BufferSize], *Head, *Tail;
inline char Getchar() {
	if(Head == Tail) {
		int l = fread(buffer, 1, BufferSize, stdin);
		Tail = (Head = buffer) + l;
	}
	return *Head++;
}
int read() {
	int x = 0, f = 1; char c = Getchar();
	while(!isdigit(c)){ if(c == ‘-‘) f = -1; c = Getchar(); }
	while(isdigit(c)){ x = x * 10 + c - ‘0‘; c = Getchar(); }
	return x * f;
}

#define maxn 100010
#define maxnode 6666666
#define oo 2147483647

int n, num[maxn<<1], cntn, head[maxn], next[maxn];
struct Point {
	int x, y;
	Point() {}
	Point(int _, int __): x(_), y(__) {}
} ps[maxn], p2[maxn];
bool cmpx(Point a, Point b) { return a.x < b.x; }
bool cmpy(Point a, Point b) { return a.y < b.y; }

int ToT, rt[maxn<<1], sumv[maxnode], lc[maxnode], rc[maxnode];
void update(int& y, int x, int l, int r, int p) {
	sumv[y = ++ToT] = sumv[x] + 1;
	if(l == r) return ;
	int mid = l + r >> 1; lc[y] = lc[x]; rc[y] = rc[x];
	if(p <= mid) update(lc[y], lc[x], l, mid, p);
	else update(rc[y], rc[x], mid + 1, r, p);
	return ;
}
void build() {
	ToT = 0;
	memset(sumv, 0, sizeof(sumv));
	memset(lc, 0, sizeof(lc));
	memset(rc, 0, sizeof(rc));
	sort(p2 + 1, p2 + n + 1, cmpy);
	for(int y = 1, i = 1; y <= cntn; y++) {
		rt[y] = rt[y-1];
		while(i <= n && p2[i].y == y) update(rt[y], rt[y], 1, cntn, p2[i].x), i++;
	}
	return ;
}
int query(int o, int l, int r, int ql, int qr) {
	if(!o) return 0;
	if(ql <= l && r <= qr) return sumv[o];
	int mid = l + r >> 1, ans = 0;
	if(ql <= mid) ans += query(lc[o], l, mid, ql, qr);
	if(qr > mid) ans += query(rc[o], mid + 1, r, ql, qr);
	return ans;
}

Point mxp[maxn<<3], mnp[maxn<<3];
void init_seg(int L, int R, int o) {
	if(L == R) mxp[o] = Point(L, -233), mnp[o] = Point(R, oo);
	else {
		int M = L + R >> 1, lc = o << 1, rc = lc | 1;
		init_seg(L, M, lc); init_seg(M+1, R, rc);
		mxp[o] = Point(L, -233); mnp[o] = Point(R, oo);
	}
	return ;
}
void modify(int L, int R, int o, Point p) {
	if(L == R) {
		if(p.y > mxp[o].y) mxp[o] = p;
		if(p.y < mnp[o].y) mnp[o] = p;
	}
	else {
		int M = L + R >> 1, lc = o << 1, rc = lc | 1;
		if(p.x <= M) modify(L, M, lc, p);
		else modify(M+1, R, rc, p);
		if(mxp[lc].y > mxp[rc].y) mxp[o] = mxp[lc]; else mxp[o] = mxp[rc];
		if(mnp[lc].y < mnp[rc].y) mnp[o] = mnp[lc]; else mnp[o] = mnp[rc];
	}
	return ;
}
void clear(int L, int R, int o, Point p) {
	if(L == R) mxp[o] = Point(L, -233), mnp[o] = Point(R, oo);
	else {
		int M = L + R >> 1, lc = o << 1, rc = lc | 1;
		if(p.x <= M) clear(L, M, lc, p);
		else clear(M+1, R, rc, p);
		mxp[o] = Point(L, -233); mnp[o] = Point(R, oo);
	}
	return ;
}
Point querymx(int L, int R, int o, int ql, int qr) {
	if(ql <= L && R <= qr) return mxp[o];
	int M = L + R >> 1, lc = o << 1, rc = lc | 1;
	Point res(L, -233), tmp;
	if(ql <= M) {
		tmp = querymx(L, M, lc, ql, qr);
		if(res.y < tmp.y) res = tmp;
	}
	if(qr > M) {
		tmp = querymx(M+1, R, rc, ql, qr);
		if(res.y < tmp.y) res = tmp;
	}
	return res;
}
Point querymn(int L, int R, int o, int ql, int qr) {
	if(ql <= L && R <= qr) return mnp[o];
	int M = L + R >> 1, lc = o << 1, rc = lc | 1;
	Point res(R, oo), tmp;
	if(ql <= M) {
		tmp = querymn(L, M, lc, ql, qr);
		if(res.y > tmp.y) res = tmp;
	}
	if(qr > M) {
		tmp = querymn(M+1, R, rc, ql, qr);
		if(res.y > tmp.y) res = tmp;
	}
	return res;
}
void build_seg(int col) {
	for(int i = head[col]; i; i = next[i]) modify(1, cntn, 1, ps[i]);
	return ;
}
void remove_seg(int col) {
	for(int i = head[col]; i; i = next[i]) clear(1, cntn, 1, ps[i]);
	return ;
}
int ans;
void Solve_down(int l, int r) {
	if(l > r) return ;
	Point tmp = querymn(1, cntn, 1, l, r);
	// matrix: x[l, r], y[1, tmp.y - 1]
	if(tmp.y < oo) ans = max(ans, query(rt[tmp.y-1], 1, cntn, l, r));
	else {
		ans = max(ans, query(rt[cntn], 1, cntn, l, r));
		return ;
	}
	Solve_down(l, tmp.x - 1); Solve_down(tmp.x + 1, r);
	return ;
}
void solve_down(int col) {
	build_seg(col);
	Solve_down(1, cntn);
	remove_seg(col);
	return ;
}
void Solve_up(int l, int r) {
	if(l > r) return ;
	Point tmp = querymx(1, cntn, 1, l, r);
	// matrix: x[l, r], y[tmp.y + 1, cntn]
//	printf("query_up: %d %d\n", tmp.x, tmp.y);
	if(tmp.y >= 0) ans = max(ans, query(rt[cntn], 1, cntn, l, r) - query(rt[tmp.y], 1, cntn, l, r));
	else {
		ans = max(ans, query(rt[cntn], 1, cntn, l, r));
		return ;
	}
//	printf("ans: %d\n", ans);
	Solve_up(l, tmp.x - 1); Solve_up(tmp.x + 1, r);
	return ;
}
void solve_up(int col) {
	build_seg(col);
	Solve_up(1, cntn);
	remove_seg(col);
	return ;
}

int main() {
	int T = read();
	while(T--) {
		memset(head, 0, sizeof(head));
		n = read(); int col = read(); cntn = 0;
		for(int i = 1; i <= n; i++) {
			int x = read(), y = read(), c = read();
			num[++cntn] = x; num[++cntn] = y;
			ps[i] = Point(x, y); next[i] = head[c]; head[c] = i;
		}
		sort(num + 1, num + cntn + 1);
		cntn = unique(num + 1, num + cntn + 1) - num - 1;
		for(int i = 1; i <= n; i++)
			ps[i].x = lower_bound(num + 1, num + cntn + 1, ps[i].x) - num,
			ps[i].y = lower_bound(num + 1, num + cntn + 1, ps[i].y) - num,
			p2[i] = ps[i];
		/*for(int i = 1; i <= n; i++) printf("%d %d\n", ps[i].x, ps[i].y);
		for(int i = 1; i <= col; i++) {
			printf("%d:", i);
			for(int j = head[i]; j; j = next[j]) printf(" (%d, %d)", ps[j].x, ps[j].y);
			putchar(‘\n‘);
		}*/
		build();
		ans = 0;
		init_seg(1, cntn, 1);
//		puts("here");
		for(int i = 1; i <= col; i++) solve_down(i);
		for(int i = 1; i <= col; i++) solve_up(i);
		printf("%d\n", ans);
	}
	
	return 0;
}