description: A robot cleaner is placed on the floor of a rectangle room, surrounded by walls. The floor consists of n rows and m columns. The rows of the floor are numbered from 1 to n from top to bottom, and columns of the floor are numbered from 1 to m from left to right. The cell on the intersection of the r-th row and the c-th column is denoted as (r,c). The initial position of the robot is (r_b, c_b).
  
  In one second, the robot moves by dr rows and dc columns, that is, after one second, the robot moves from the cell (r, c) to (r + dr, c + dc). Initially dr = 1, dc = 1. If there is a vertical wall (the left or the right walls) in the movement direction, dc is reflected before the movement, so the new value of dc is -dc. And if there is a horizontal wall (the upper or lower walls), dr is reflected before the movement, so the new value of dr is -dr.
  
  Each second (including the moment before the robot starts moving), the robot cleans every cell lying in the same row or the same column as its position. There is only one dirty cell at (r_d, c_d). The job of the robot is to clean that dirty cell.
  
  <image> Illustration for the first example. The blue arc is the robot. The red star is the target dirty cell. Each second the robot cleans a row and a column, denoted by yellow stripes.
  
  Given the floor size n and m, the robot's initial position (r_b, c_b) and the dirty cell's position (r_d, c_d), find the time for the robot to do its job.
  
  Input
  
  Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
  
  A test case consists of only one line, containing six integers n, m, r_b, c_b, r_d, and c_d (1 ≤ n, m ≤ 100, 1 ≤ r_b, r_d ≤ n, 1 ≤ c_b, c_d ≤ m) — the sizes of the room, the initial position of the robot and the position of the dirt cell.
  
  Output
  
  For each test case, print an integer — the time for the robot to clean the dirty cell. We can show that the robot always cleans the dirty cell eventually.
  
  Example
  
  Input
  
  
  5
  10 10 6 1 2 8
  10 10 9 9 1 1
  9 8 5 6 2 1
  6 9 2 2 5 8
  2 2 1 1 2 1
  
  
  Output
  
  
  7
  10
  9
  3
  0
  
  Note
  
  In the first example, the floor has the size of 10× 10. The initial position of the robot is (6, 1) and the position of the dirty cell is (2, 8). See the illustration of this example in the problem statement.
  
  In the second example, the floor is the same, but the initial position of the robot is now (9, 9), and the position of the dirty cell is (1, 1). In this example, the robot went straight to the dirty cell and clean it. 
  
  <image>
  
  In the third example, the floor has the size 9 × 8. The initial position of the robot is (5, 6), and the position of the dirty cell is (2, 1). 
  
  <image>
  
  In the fourth example, the floor has the size 6 × 9. The initial position of the robot is (2, 2) and the position of the dirty cell is (5, 8). 
  
  <image>
  
  In the last example, the robot was already standing in the same column as the dirty cell, so it can clean the cell right away. 
  
  <image>
solution: #include <bits/stdc++.h>
  using namespace std;
  constexpr int bits(int x) { return x == 0 ? 0 : 31 - __builtin_clz(x); }
  using ld = long double;
  using ll = long long;
  using vi = vector<int>;
  using vvi = vector<vi>;
  using vll = vector<ll>;
  using vvll = vector<vll>;
  using vb = vector<bool>;
  using vvb = vector<vb>;
  using vd = vector<double>;
  using vs = vector<string>;
  using pii = pair<int, int>;
  using pll = pair<ll, ll>;
  using pdd = pair<double, double>;
  using vpii = vector<pii>;
  using vpll = vector<pll>;
  using vpdd = vector<pdd>;
  template <typename T>
  void ckmin(T &a, const T &b) {
    a = min(a, b);
  }
  template <typename T>
  void ckmax(T &a, const T &b) {
    a = max(a, b);
  }
  mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
  namespace __input {
  template <class T1, class T2>
  void re(pair<T1, T2> &p);
  template <class T>
  void re(vector<T> &a);
  template <class T, size_t SZ>
  void re(array<T, SZ> &a);
  template <class T>
  void re(T &x) {
    cin >> x;
  }
  void re(double &x) {
    string t;
    re(t);
    x = stod(t);
  }
  template <class Arg, class... Args>
  void re(Arg &first, Args &...rest) {
    re(first);
    re(rest...);
  }
  template <class T1, class T2>
  void re(pair<T1, T2> &p) {
    re(p.f, p.s);
  }
  template <class T>
  void re(vector<T> &a) {
    for (int i = 0; i < int((a).size()); i++) re(a[i]);
  }
  template <class T, size_t SZ>
  void re(array<T, SZ> &a) {
    for (int i = 0; i < SZ; i++) re(a[i]);
  }
  }  // namespace __input
  using namespace __input;
  namespace __output {
  template <typename T>
  struct is_outputtable {
    template <typename C>
    static constexpr decltype(declval<ostream &>() << declval<const C &>(),
                              bool())
    test(int) {
      return true;
    }
    template <typename C>
    static constexpr bool test(...) {
      return false;
    }
    static constexpr bool value = test<T>(int());
  };
  template <
      class T, typename V = decltype(declval<const T &>().begin()),
      typename second = typename enable_if<!is_outputtable<T>::value, bool>::type>
  void pr(const T &x);
  template <class T,
            typename V = decltype(declval<ostream &>() << declval<const T &>())>
  void pr(const T &x) {
    cout << x;
  }
  template <class T1, class T2>
  void pr(const pair<T1, T2> &x);
  template <class Arg, class... Args>
  void pr(const Arg &first, const Args &...rest) {
    pr(first);
    pr(rest...);
  }
  template <class T, bool pretty = true>
  void prContain(const T &x) {
    if (pretty) pr("{");
    bool fst = 1;
    for (const auto &a : x) pr(!fst ? pretty ? ", " : " " : "", a), fst = 0;
    if (pretty) pr("}");
  }
  template <class T>
  void pc(const T &x) {
    prContain<T, false>(x);
    pr("\n");
  }
  template <class T1, class T2>
  void pr(const pair<T1, T2> &x) {
    pr("{", x.f, ", ", x.s, "}");
  }
  template <class T, typename V, typename second>
  void pr(const T &x) {
    prContain(x);
  }
  void ps() { pr("\n"); }
  template <class Arg>
  void ps(const Arg &first) {
    pr(first);
    ps();
  }
  template <class Arg, class... Args>
  void ps(const Arg &first, const Args &...rest) {
    pr(first, " ");
    ps(rest...);
  }
  }  // namespace __output
  using namespace __output;
  namespace __algorithm {
  template <typename T>
  void dedup(vector<T> &v) {
    sort((v).begin(), (v).end());
    v.erase(unique((v).begin(), (v).end()), v.end());
  }
  template <typename T>
  typename vector<T>::const_iterator find(const vector<T> &v, const T &x) {
    auto it = lower_bound((v).begin(), (v).end(), x);
    return it != v.end() && *it == x ? it : v.end();
  }
  template <typename T>
  size_t index(const vector<T> &v, const T &x) {
    auto it = find(v, x);
    assert(it != v.end() && *it == x);
    return it - v.begin();
  }
  template <typename I>
  struct _reversed_struct {
    I &v_;
    explicit _reversed_struct(I &v) : v_{v} {}
    typename I::reverse_iterator begin() const { return v_.rbegin(); }
    typename I::reverse_iterator end() const { return v_.rend(); }
  };
  template <typename I>
  _reversed_struct<I> reversed(I &v) {
    return _reversed_struct<I>(v);
  }
  }  // namespace __algorithm
  using namespace __algorithm;
  namespace __io {
  void setIO() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout << setprecision(15);
  }
  }  // namespace __io
  using namespace __io;
  template <typename T_vector>
  void outvec(const T_vector &v, bool add_one = false, int start = -1,
              int end = -1) {
    if (start < 0) start = 0;
    if (end < 0) end = int(v.size());
    for (int i = start; i < end; i++)
      cout << v[i] + (add_one ? 1 : 0) << (i < end - 1 ? ' ' : '\n');
  }
  const ld pi = 3.14159265358979323846;
  const ll mod = 1000000007;
  const ll N = 7e5;
  ll n, m, k, l, r, x, y, z;
  const ll template_array_size = 1e6 + 1851;
  ll ar[template_array_size];
  ll br[template_array_size];
  ll cr[template_array_size];
  string s, t;
  void solve() {
    int n, m, rb, cb, rd, cd;
    re(n, m, rb, cb, rd, cd);
    int steps = 0;
    int dx = 1, dy = 1;
    while (1) {
      if (rb == rd || cb == cd) {
        ps(steps);
        return;
      }
      steps++;
      if (rb + dx > n || rb + dx <= 0) {
        dx = -dx;
      }
      if (cb + dy > m || cb + dy <= 0) {
        dy = -dy;
      }
      rb += dx;
      cb += dy;
    }
  }
  int32_t main() {
    setIO();
    int t;
    cin >> t;
    for (int i = 0; i < t; i++) {
      solve();
    }
    return 0;
  }
  