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.
#include <bits/stdc++.h>
using namespace std;
using int64 = long long;
int64 m;
int64 p2[55];
pair<int64, int64> bounds(int64 x0, int terms) {
  assert(terms > 0);
  return make_pair(p2[terms - 1] * (x0 + 1), p2[terms - 1] * (x0 + m));
}
int main(int argc, char** argv) {
  const bool debug = (argc > 1);
  p2[0] = 1;
  for (int i = 1; i < 55; ++i) p2[i] = 2 * p2[i - 1];
  int q;
  scanf("%d", &q);
  while (q--) {
    int64 a, target;
    scanf("%lld%lld%lld", &a, &target, &m);
    if (debug) printf("x0=%lld, x[n-1]=%lld, m=%lld\n", a, target, m);
    if (a == target) {
      printf("1 %lld\n", a);
      continue;
    }
    int terms = 0;
    bool possible = true;
    int64 lo[55], sumlo[55];
    int64 hi[55], sumhi[55];
    sumlo[0] = sumhi[0] = lo[0] = hi[0] = a;
    for (int i = 1; i < 50; ++i) {
      lo[i] = sumlo[i - 1] + 1;
      hi[i] = sumhi[i - 1] + m;
      if (debug) {
        printf(">>>>> %lld %lld\n", lo[i], hi[i]);
      }
      sumlo[i] = sumlo[i - 1] + lo[i];
      sumhi[i] = sumhi[i - 1] + hi[i];
      if (hi[i] >= target) {
        if (lo[i] > target) {
          possible = false;
        }
        terms = i + 1;
        break;
      }
    }
    if (!possible) {
      printf("-1\n");
      continue;
    }
    assert(terms > 1);
    assert(terms <= 50);
    if (m == 1) {
      printf("%d", terms);
      for (int i = 0; i < terms; ++i) {
        printf(" %lld", lo[i]);
      }
      printf("\n");
      continue;
    }
    int64 x[55];
    if (debug) printf("terms=%d\n", terms);
    auto res = bounds(a, terms - 1);
    if (debug) printf("init:: lo=%lld, hi=%lld\n", res.first, res.second);
    int64 runsumx = x[0] = a;
    for (int i = 1; i < terms - 1; ++i) {
      int terms_after = terms - i - 1;
      assert(terms_after >= 1);
      if (debug)
        printf("\ni=%d ^=%d p2=%lld\n", i, terms_after - 1,
               p2[terms_after - 1]);
      if (debug)
        printf("div=%lld, 2rx=%lld\n", (target / p2[terms_after - 1]),
               2 * runsumx);
      int64 r = ((target + p2[terms_after - 1] - 1) / p2[terms_after - 1]) -
                2 * runsumx - m;
      if (debug) printf("r=%lld\n", r);
      if (r < 1) r = 1;
      if (r > m) r = m;
      assert(r >= 1);
      assert(r <= m);
      x[i] = runsumx + r;
      runsumx += x[i];
      if (debug) printf("x[%d]=%lld, runsumx=%lld\n", i, x[i], runsumx);
    }
    assert(target - runsumx >= 1);
    assert(target - runsumx <= m);
    x[terms - 1] = target;
    printf("%d", terms);
    for (int i = 0; i < terms; ++i) {
      printf(" %lld", x[i]);
    }
    printf("\n");
  }
  return 0;
}
