Problem: Given a positive integer m, we say that a sequence x_1, x_2, ..., x_n of positive integers is m-cute if for every index i such that 2 ≤ i ≤ n it holds that x_i = x_{i - 1} + x_{i - 2} + ... + x_1 + r_i for some positive integer r_i satisfying 1 ≤ r_i ≤ m.

You will be given q queries consisting of three positive integers a, b and m. For each query you must determine whether or not there exists an m-cute sequence whose first term is a and whose last term is b. If such a sequence exists, you must additionally find an example of it.

Input

The first line contains an integer number q (1 ≤ q ≤ 10^3) — the number of queries.

Each of the following q lines contains three integers a, b, and m (1 ≤ a, b, m ≤ 10^{14}, a ≤ b), describing a single query.

Output

For each query, if no m-cute sequence whose first term is a and whose last term is b exists, print -1.

Otherwise print an integer k (1 ≤ k ≤ 50), followed by k integers x_1, x_2, ..., x_k (1 ≤ x_i ≤ 10^{14}). These integers must satisfy x_1 = a, x_k = b, and that the sequence x_1, x_2, ..., x_k is m-cute.

It can be shown that under the problem constraints, for each query either no m-cute sequence exists, or there exists one with at most 50 terms.

If there are multiple possible sequences, you may print any of them.

Example

Input


2
5 26 2
3 9 1


Output


4 5 6 13 26
-1

Note

Consider the sample. In the first query, the sequence 5, 6, 13, 26 is valid since 6 = 5 + \bf{\color{blue} 1}, 13 = 6 + 5 + {\bf\color{blue} 2} and 26 = 13 + 6 + 5 + {\bf\color{blue} 2} have the bold values all between 1 and 2, so the sequence is 2-cute. Other valid sequences, such as 5, 7, 13, 26 are also accepted.

In the second query, the only possible 1-cute sequence starting at 3 is 3, 4, 8, 16, ..., which does not contain 9.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
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.first, p.second);
}
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 S = 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.first, ", ", x.second, "}");
}
template <class T, typename V, typename S>
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;
void solve() {
  ll a, b, m;
  re(a, b, m);
  if (a > b) {
    ps(-1);
    return;
  }
  if (a == b) {
    ps(1, a);
    return;
  }
  vll powers{1};
  ll v = a;
  for (; v < b; v = 2 * v + 1) {
    powers.push_back(powers.back() * 2);
  }
  powers.pop_back();
  powers.pop_back();
  v = (v - 1) / 2;
  ll d = max(0LL, b - (v + m));
  if (b - (v + 1) < 0) {
    ps(-1);
    return;
  }
  vll coeff;
  for (ll p : reversed(powers)) {
    ll c = min(m - 1, d / p);
    coeff.push_back(c);
    d -= p * c;
  }
  if (d > 0) {
    ps(-1);
    return;
  }
  vll second{0, a};
  for (ll c : coeff) {
    second.push_back(second.back() * 2 + 1 + c);
  }
  vll ans;
  for (int i = 0; i + 1 < int((second).size()); i++) {
    ans.push_back(second[i + 1] - second[i]);
  }
  ans.push_back(b);
  pr(int((ans).size()), " ");
  pc(ans);
}
int main() {
  setIO();
  int T;
  re(T);
  for (int t = 0; t < T; t++) solve();
  return 0;
}