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;
void soln() {
  long long int a, b, m;
  cin >> a >> b >> m;
  if (a == b) {
    cout << 1 << " " << a << endl;
    return;
  }
  if (a + m >= b) {
    cout << 2 << " " << a << " " << b << endl;
    return;
  }
  long long int ans = a;
  long long int count = 1;
  long long int sum = a;
  while (1) {
    ans += m;
    count++;
    if (ans >= b) break;
    sum += ans;
    ans = sum;
  }
  vector<long long int> v1;
  long long int l = count - 1;
  long long int p = 1;
  for (int i = 1; i < l; i++) {
    p *= 2;
  }
  long long int x = b - p * a;
  long long int m1 = x / p;
  if (m1 < 1) {
    cout << -1 << endl;
    return;
  }
  v1.push_back(a);
  ans = a;
  sum = a;
  for (int i = 1; i <= l; i++) {
    ans += m1;
    v1.push_back(ans);
    sum += ans;
    ans = sum;
  }
  long long int limit = b - v1[v1.size() - 1];
  vector<long long int> v2;
  while (limit > 0) {
    v2.push_back(limit % 2);
    limit /= 2;
  }
  for (int i = v2.size(); i < v1.size() - 1; i++) {
    v2.push_back(0);
  }
  reverse(v2.begin(), v2.end());
  v2.push_back(0);
  long long int cnt = 0;
  long long int s = 0, c = 0, prev = 0;
  for (int i = 0; i < v1.size(); i++) {
    v1[i] += (2 * s);
    v1[i] += prev;
    s *= 2;
    if (prev == 1) {
      s++;
      prev = 0;
    }
    if (v2[i] == 1) prev = 1;
    v1[i] += prev;
  }
  cout << v1.size() << " ";
  for (int i = 0; i < v1.size(); i++) {
    cout << v1[i] << " ";
  }
  cout << endl;
}
int main() {
  long long int q;
  cin >> q;
  while (q--) {
    soln();
  }
}
