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>
#pragma GCC optimize("Ofast")
using namespace std;
mt19937 rng32(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rng64(chrono::steady_clock::now().time_since_epoch().count());
long long a, b, m;
void Input() { cin >> a >> b >> m; }
void Solve() {
  if (a == b) {
    cout << "1 " << a << '\n';
    return;
  }
  vector<long long> Min, Max;
  Min.push_back(a);
  Max.push_back(a);
  int siz = 1;
  for (; siz <= 50; siz++) {
    if (siz == 50) {
      siz++;
      break;
    }
    long long Mn = 1, Mx = m;
    for (int x = 0; x < siz; x++) {
      Mn += Min[x];
      Mx += Max[x];
    }
    Min.push_back(Mn);
    Max.push_back(Mx);
    if (Mn > b) {
      cout << "-1\n";
      return;
    }
    if (Mn <= b && b <= Mx) {
      siz++;
      break;
    }
  }
  if (siz > 50) {
    cout << "-1\n";
    return;
  }
  long long curr = Min.back();
  vector<long long> Weight(siz, 1), coef(siz, 1);
  for (int i = 2; i < siz; i++) Weight[i] = Weight[i - 1] * 2;
  reverse(Weight.begin(), Weight.end());
  for (int i = 1; i < siz; i++) {
    long long MaxAllowed = min(m - 1, (b - curr) / Weight[i]);
    coef[i] += MaxAllowed;
    curr += MaxAllowed * Weight[i];
  }
  vector<long long> arr(siz, 0);
  arr[0] = a;
  for (int i = 1; i < siz; i++) {
    for (int j = 0; j < i; j++) arr[i] += arr[j];
    arr[i] += coef[i];
  }
  cout << siz;
  for (auto z : arr) cout << " " << z;
  cout << '\n';
}
int main(int argc, char* argv[]) {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  int T;
  cin >> T;
  while (T--) {
    Input();
    Solve();
  }
  return 0;
}