1
  2
  3
  4
  5
  6
  7
  8
  9
 10
 11
 12
 13
 14
 15
 16
 17
 18
 19
 20
 21
 22
 23
 24
 25
 26
 27
 28
 29
 30
 31
 32
 33
 34
 35
 36
 37
 38
 39
 40
 41
 42
 43
 44
 45
 46
 47
 48
 49
 50
 51
 52
 53
 54
 55
 56
 57
 58
 59
 60
 61
 62
 63
 64
 65
 66
 67
 68
 69
 70
 71
 72
 73
 74
 75
 76
 77
 78
 79
 80
 81
 82
 83
 84
 85
 86
 87
 88
 89
 90
 91
 92
 93
 94
 95
 96
 97
 98
 99
100
101
102
103
#include <stdio.h>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <stack>

using namespace std;

typedef vector<int> VI;
typedef long long LL;
typedef pair<int,int> para;

#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 FOREACH(i, c) for(VAR(i, (c).begin()); i != (c).end(); ++i)
#define PB push_back
#define ST first
#define ND second
#define MP make_pair

struct bryka {
  int h;
  int x1;
  int x2;
};

bool compare(const bryka &a, const bryka &b) {
  return a.h > b.h;
}


int main() {
  int t, n, w, d1, d2, d3, d4, h, x1, x2;
  bryka b;

  scanf("%d", &t);

  REP(i,t) {
    vector<bryka> ord;
    stack<VI> s;
    VI all, left, right;
int m=0;

    scanf("%d%d", &n, &w);
    ord.reserve(n);

    REP(j,n) {
      scanf("%d%d%d%d", &d1, &d2, &d3, &d4);
      b.h=abs(d2-d4);
      b.x1=d1;
      ord.PB(b);
    }

    REP(j,n) {
      scanf("%d%d%d%d", &d1, &d2, &d3, &d4);
      ord[j].x2=d1;
    }

    if (n==1) {cout<<"TAK\n"; continue;}

    sort(ALL(ord), compare);

    REP(j,n) all.PB(j);
    s.push(all);

    while (!s.empty()) {
      bool l1, l2;
      all=s.top();
      s.pop();
      b=ord[all[0]];
      h=b.h;
      x1=b.x1;
      x2=b.x2;
m++;
      FOR(j,1,all.size()-1) {
        int k=all[j];
        b=ord[k];
        if (h+b.h<=w) break;

        l1= b.x1<x1; l2= b.x2<x2;
        if (l1!=l2) {cout<<"NIE\n"; goto NIE;}

        if (l1) left.PB(k); else right.PB(k);
      }

      if (left.size()>1) s.push(left);
      if (right.size()>1) s.push(right);
      all.clear(); left.clear(); right.clear();


    }
    printf("%s","TAK\n");

  NIE: ;
  }

  return 0;
}