/**
 *  Railroad Conflict (Problem D, Practice Contest for Japan Domestic, 2006-06-18)
 *  by Kenji Inoue, 2007-01-07
 **/

#include <iostream>
#include <vector>
#include <complex>

using namespace std;

typedef complex<double> P;

#define EPS (1e-9)

double cross(P a, P b) {
  return (a.real() * b.imag() - a.imag() * b.real());
}

int is_intersected(P a1, P a2, P b1, P b2) {
  return ( ( cross(a2-a1, b1-a1) * cross(a2-a1, b2-a1) < EPS ) &&
           ( cross(b2-b1, a1-b1) * cross(b2-b1, a2-b1) < EPS ) );
}

P s[101], t[101];
int c[101], l[101];

struct intersection {
  double t; // parameter t
  int n;    // index of intersected line

  bool operator<(const intersection &a) const {
    return (t < a.t);
  }
};

int main() {
  int N;
  cin >> N;
  while (N--) {
    P a, b;
    cin >> a.real() >> a.imag();
    cin >> b.real() >> b.imag();

    int n;
    cin >> n;
    for (int i=0; i<n; i++) {
      cin >> s[i].real() >> s[i].imag();
      cin >> t[i].real() >> t[i].imag();
      cin >> c[i] >> l[i];
    }

    vector<intersection> is;
    for (int i=0; i<n; i++) {
      if (!is_intersected(a, b, s[i], t[i])) { continue; }
      if (abs(cross(b-a, t[i]-s[i])) < EPS) { continue; }

      P x = t[i] - s[i];
      double d1 = abs(cross(x, a-s[i]));
      double d2 = abs(cross(x, b-s[i]));
      double t = d1 / (d1 + d2);

      intersection tmp = {t, i};
      is.push_back(tmp);
    }

    sort(is.begin(), is.end());

    int curr; // current status: up (=1) or down (=0)
    if (is.size() > 0) {
      int idx = is[0].n;
      curr = (c[idx] == 1 ? l[idx] : !l[idx]);
    }

    int cnt = 0;
    for (int i=1; i<is.size(); i++) {
      int idx = is[i].n;
      if ( (c[idx] == 1 && curr != l[idx]) ||
           (c[idx] == 0 && curr == l[idx]) ) {
        curr = !curr;
        cnt++;
      }
    }

    cout << cnt << endl;
  }

  return 0;
}