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;
bool slove(int64 a, int64 b, int64 m, int k) {
  vector<int64> p(k);
  p[k - 2] = p[k - 1] = 1;
  for (int i = k - 3; ~i; i--) p[i] = p[i + 1] + p[i + 1];
  if (b / p[0] < a) return false;
  b -= a * p[0];
  for (int i = 1; i < k; i++) {
    if ((b -= p[i]) < 0) return false;
  }
  vector<int64> ans(k, 1);
  ans[0] = a;
  for (int i = 1; i < k; i++) {
    int64 t = min(b / p[i], m - 1);
    b -= t * p[i], ans[i] += t;
  }
  if (b != 0) return false;
  int64 sum = 0;
  for (int i = 0; i < k; i++) {
    ans[i] += sum;
    sum += ans[i];
  }
  printf("%d\n", (int)ans.size());
  for (int i = 0; i < k; i++) printf("%lld%c", ans[i], " \n"[i + 1 == k]);
  return true;
}
int main() {
  int T;
  for (scanf("%d", &T); T--;) {
    int64 a, b, m;
    scanf("%lld%lld%lld", &a, &b, &m);
    if (a == b) {
      printf("1 %lld\n", a);
      continue;
    }
    bool ok = false;
    for (int k = 2; k <= 50 && !ok; k++) ok |= slove(a, b, m, k);
    if (!ok) puts("-1");
  }
  return 0;
}
