#include <bits/stdc++.h>
#define all(x) begin(x), end(x)
#define rep(i, l, r) for (int i = (l), i##end = (r); i <= i##end; ++i)
#define per(i, l, r) for (int i = (l), i##end = (r); i >= i##end; --i)
#ifdef memset0
#define log(...) fprintf(stderr, __VA_ARGS__)
#else
#define log(...) (void(0))
#define endl '\n'
#endif
using namespace std;
using ll = long long;
using lf = long double;
using ull = unsigned long long;

const int N = 1e2 + 9;
const int mov[8][3] = {
    {1, 0, 0}, {-1, 0, 0}, {0, 1, 0}, {0, -1, 0}, {0, 0, 1}, {0, 0, -1},
};
int T, n, m, l, r, s[N][N][N], dis[N][N][N];
struct atom {
  int x, y, z;
} q[N * N * N], a, b;

int solve() {
  if (s[a.x][a.y][a.z] || s[b.x][b.y][b.z]) return -1;
  q[l = r = 0] = a;
  dis[a.x][a.y][a.z] = 1;
  while (l <= r) {
    auto u = q[l++];
    // log("> %d %d %d >> dis=%d\n", u.x, u.y, u.z, dis[u.x][u.y][u.z]);
    for (int i = 0; i < 8; i++) {
      auto v = u;
      v.x += mov[i][0];
      v.y += mov[i][1];
      v.z += mov[i][2];
      if (1 <= v.x && v.x <= n && 1 <= v.y && v.y <= n && 1 <= v.z && v.z <= n && !s[v.x][v.y][v.z] && !dis[v.x][v.y][v.z]) {
        dis[v.x][v.y][v.z] = dis[u.x][u.y][u.z] + 1;
        q[++r] = v;
      }
    }
  }
  return dis[b.x][b.y][b.z] - 1;
}

int main() {
#ifdef memset0
  freopen("H.in", "r", stdin);
#endif
  cin.tie(0)->sync_with_stdio(0);
  cin >> T;
  while (T--) {
    cin >> n >> m;
    rep(i, 0, n) rep(j, 0, n) rep(k, 0, n) { s[i][j][k] = dis[i][j][k] = 0; }
    rep(k, 1, m) {
      int x0, x1, y0, y1, z0, z1;
      cin >> x0 >> y0 >> z0 >> x1 >> y1 >> z1;
      s[x0][y0][z0]++;
      s[x1 + 1][y0][z0]--;
      s[x0][y1 + 1][z0]--;
      s[x0][y0][z1 + 1]--;
      s[x1 + 1][y1 + 1][z0]++;
      s[x1 + 1][y0][z1 + 1]++;
      s[x0][y1 + 1][z1 + 1]++;
      s[x1 + 1][y1 + 1][z1 + 1]--;
    }
    rep(i, 0, n) rep(j, 0, n) rep(k, 0, n) {
      if (i) s[i][j][k] += s[i - 1][j][k];
      if (j) s[i][j][k] += s[i][j - 1][k];
      if (k) s[i][j][k] += s[i][j][k - 1];
      if (i && j) s[i][j][k] -= s[i - 1][j - 1][k];
      if (i && k) s[i][j][k] -= s[i - 1][j][k - 1];
      if (j && k) s[i][j][k] -= s[i][j - 1][k - 1];
      if (i && j && k) s[i][j][k] += s[i - 1][j - 1][k - 1];
    }
    // rep(i, 1, n) rep(j, 1, n) rep(k, 1, n) log("a[%d][%d][%d] = %d\n", i, j, k, s[i][j][k]);
    cin >> a.x >> a.y >> a.z >> b.x >> b.y >> b.z;
    cout << solve() << endl;
  }
}