#include <bits/stdc++.h>
using namespace std;
long long dp[55];
void p_2() {
  dp[0] = 1;
  for (int i = 1; i <= 50; i++) dp[i] = dp[i - 1] * 2;
}
int main() {
  ios_base ::sync_with_stdio(false);
  int testcase;
  cin >> testcase;
  long long a, b, m;
  vector<long long> vec;
  long long last;
  long long dif;
  long long add[55];
  int x;
  long long temp;
  p_2();
  while (testcase--) {
    cin >> a >> b >> m;
    if (a == b) {
      cout << 1 << ' ' << a << endl;
      continue;
    } else if (b - a <= m) {
      cout << 2 << ' ' << a << ' ' << b << endl;
      continue;
    }
    vec.clear();
    memset(add, 0, sizeof add);
    vec.push_back(a);
    vec.push_back(a + 1);
    last = a + 1;
    while (last <= b) {
      last *= 2;
      if (last > b) break;
      vec.push_back(last);
    }
    dif = b - last / 2;
    for (int i = 1; i < vec.size() - 1; i++) {
      x = vec.size() - i - 2;
      long long temp = dp[x];
      add[i] += min((dif / temp), m - 1);
      if (min(dif / temp, m - 1) != 0) dif -= (min(dif / temp, m - 1) * temp);
      if (dif == 0) break;
    }
    if (dif != 0) {
      if (dif <= m - 1)
        add[vec.size() - 1] = dif;
      else {
        cout << -1 << endl;
        continue;
      }
    }
    cout << vec.size() << ' ';
    cout << vec[0] << ' ' << vec[1] + add[1] << ' ';
    last = add[1];
    for (int i = 2; i < vec.size(); i++) {
      cout << vec[i] + last + add[i] << ' ';
      last *= 2;
      last += add[i];
    }
    cout << endl;
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long dp[55];
void p_2() {
  dp[0] = 1;
  for (int i = 1; i <= 50; i++) dp[i] = dp[i - 1] * 2;
}
int main() {
  ios_base ::sync_with_stdio(false);
  int testcase;
  cin >> testcase;
  long long a, b, m;
  vector<long long> vec;
  long long last;
  long long dif;
  long long add[55];
  int x;
  long long temp;
  p_2();
  while (testcase--) {
    cin >> a >> b >> m;
    if (a == b) {
      cout << 1 << ' ' << a << endl;
      continue;
    } else if (b - a <= m) {
      cout << 2 << ' ' << a << ' ' << b << endl;
      continue;
    }
    vec.clear();
    memset(add, 0, sizeof add);
    vec.push_back(a);
    vec.push_back(a + 1);
    last = a + 1;
    while (last <= b) {
      last *= 2;
      if (last > b) break;
      vec.push_back(last);
    }
    dif = b - last / 2;
    for (int i = 1; i < vec.size() - 1; i++) {
      x = vec.size() - i - 2;
      long long temp = dp[x];
      add[i] += min((dif / temp), m - 1);
      if (min(dif / temp, m - 1) != 0) dif -= (min(dif / temp, m - 1) * temp);
      if (dif == 0) break;
    }
    if (dif != 0) {
      if (dif <= m - 1)
        add[vec.size() - 1] = dif;
      else {
        cout << -1 << endl;
        continue;
      }
    }
    cout << vec.size() << ' ';
    cout << vec[0] << ' ' << vec[1] + add[1] << ' ';
    last = add[1];
    for (int i = 2; i < vec.size(); i++) {
      cout << vec[i] + last + add[i] << ' ';
      last *= 2;
      last += add[i];
    }
    cout << endl;
  }
  return 0;
}

```