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;
long long a, b, m, bit[100] = {}, bit2[100] = {};
int n, MAX_LEN, MAX_BIT;
void rev(int len) {
  for (int i = len; i >= 1; i--) {
    bit2[len - i + 1] = bit[i];
  }
  bit2[len + 1] = floorl(bit[0] / 2.0);
  bit2[len + 2] = ceill(bit[0] / 2.0);
}
void PRINT(int len) {
  rev(len - 3);
  cout << len << " " << a;
  long long tt = (long long)(a + bit2[1]);
  for (int j = 2; j <= len; j++) {
    cout << " " << tt;
    tt = (long long)(tt * 2 - bit2[j - 1] + bit2[j]);
  }
  cout << endl;
}
void DO() {
  if (a == b) {
    cout << 1 << " " << a << endl;
    return;
  } else if (a + m >= b) {
    cout << 2 << " " << a << " " << b << endl;
    return;
  }
  MAX_LEN = (int)(logl(b / (a + 1)) / log(2)) + 2;
  bit[0] = 2;
  for (int i = 1; i <= MAX_LEN - 3; i++) bit[i] = 1;
  long long val = pow(2, MAX_LEN - 2) * (a + 1);
  long long diff = b - val;
  if (diff == 0) {
    PRINT(MAX_LEN);
    return;
  }
  MAX_BIT = (int)(logl(diff) / log(2));
  MAX_BIT = fmin(MAX_LEN - 3, MAX_BIT);
  for (int i = MAX_BIT; i >= 0; i--) {
    long long coeff = diff / pow(2, i);
    diff -= coeff * pow(2, i);
    if (i == 0) {
      bit[i] += coeff;
    } else {
      if (bit[i] + coeff > m) {
        bit[i - 1] += 2 * (bit[i] + coeff - m);
        bit[i] = m;
      } else
        bit[i] += coeff;
    }
  }
  if (bit[0] > 2 * m || diff > 0)
    cout << "-1\n";
  else
    PRINT(MAX_LEN);
  return;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a >> b >> m;
    DO();
  }
  return 0;
}
