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/*
W rozwiązaniu wykorzystałem implementację drzewa przedziałowego ze strony:
http://was.zaa.mimuw.edu.pl/?q=node/9
*/

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

typedef long long int LL;
typedef long double LD;
typedef vector<int> VI;
typedef vector<LL> VLL;
typedef vector<LD> VLD;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef vector<PII> VPII;

const int INF = 1000000001;
const LD EPS = 10e-9;

#define FOR(x, b, e) for(int x = b; x <= (e); ++x)
#define FORD(x, b, e) for(int x = b; x >= (e); --x)
#define REP(x, n) for(int x = 0; x < (n); ++x)
#define VAR(v, n) __typeof(n) v = (n)
#define ALL(c) (c).begin(), (c).end()
#define SIZE(x) ((int)(x).size())
#define FORE(i, c) for(VAR(i, (c).begin()); i != (c).end(); ++i)
#define MP make_pair
#define PB push_back
#define ST first
#define ND second
#define abs(a) ((a)<0 ? -(a) : (a))
#define max(a, b) ((a) > (b) ? (a) : (b))
#define min(a, b) ((a) < (b) ? (a) : (b))

const int MAX_N = 60005;
const int MAX_M = 65536;

int tree[2*MAX_M];

void insert(int x, int val) 
{ 
	int v = MAX_M + x;
	tree[v] = val;
	while (v != 1) 
	{
		v /= 2;
		tree[v] = max(tree[2*v], tree[2*v+1]);
	}
}

int query(int a, int b) 
{
	if(a > b) return 0;
	
	int va = MAX_M + a, vb = MAX_M + b;
	int wyn = tree[va];
	if (va != vb) wyn = max(wyn, tree[vb]);

	while (va / 2 != vb / 2) 
	{
		if (va % 2 == 0) wyn = max(wyn, tree[va+1]); 
		if (vb % 2 == 1) wyn = max(tree[vb-1], wyn); 
		va /= 2; vb /= 2;
	}
	return wyn;
}

struct car
{
	int w, x2, id;
	
	car(int _w = 0, int _x2 = 0, int _id = 0) : w(_w), x2(_x2), id(_id) {};	
}carsIn[MAX_N], carsInSort[MAX_N], carsOut[MAX_N];

bool operator<(const car &a, const car &b)
{
	if(a.x2 == b.x2)
	{
		return a.id < b.id;
	}
	return a.x2 < b.x2;
}

int t, n, x1_, x2_, y1_, y2_, w;

int main()
{
	scanf("%d", &t);
	{
		while(t--)
		{
			scanf("%d %d", &n, &w);
			REP(i, n)
			{
				scanf("%d %d %d %d", &x1_, &y1_, &x2_, &y2_);
				carsInSort[i] = carsIn[i] = car(abs(y2_-y1_), max(x1_, x2_), i);	
			}	
			
			sort(carsInSort, carsInSort+n);
			REP(i, n)
			{
				carsInSort[i].x2 = carsIn[carsInSort[i].id].x2 = i;
				insert(i, carsInSort[i].w);
				//cout << "insert w: " << carsInSort[i].w << endl;
			}
			
			REP(i, n)
			{
				scanf("%d %d %d %d", &x1_, &y1_, &x2_, &y2_);
				carsOut[i] = car(abs(y2_-y1_), max(x1_, x2_), i);
			}	
			
			bool ok = 1;
			sort(carsOut, carsOut+n);
			REP(i, n)
			{
				car carOrig = carsIn[carsOut[i].id];
				int q = query(0, carOrig.x2-1);
				//cout << "id: " << carOrig.id << ", q: " << q << ", right: " << carOrig.x2-1 << endl;
				if(q > w-carOrig.w)
				{
					ok = 0;
					break;
				}
				
				insert(carOrig.x2, 0);
			}
			
			printf((ok ? "TAK\n" : "NIE\n"));
			memset(tree, 0, (2*MAX_M-1)*sizeof(int));			
		}
	}	
	//system("pause");
	return 0;
}