#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vi = vector<int>;
using pii = pair<int, int>;
const int MAXN = 1e5 + 10;
template <typename T>
istream &operator>>(istream &is, vector<T> &v) {
  for (auto i = begin(v); i != end(v); i++) is >> *i;
  return is;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  for (auto i = begin(v); i != end(v); i++)
    os << *i << (i == end(v) - 1 ? "\n" : " ");
  return os;
}
signed main() {
  int t;
  cin >> t;
  while (t--) {
    ll a, b, m;
    cin >> a >> b >> m;
    if (a >= b) {
      if (a > b) cout << "-1" << '\n';
      if (a == b) cout << "1 " << a << '\n';
      continue;
    }
    int n;
    for (n = 1; n <= 50; ++n) {
      ll l = (1ll << (n - 1)) * (a + 1), r = (1ll << (n - 1)) * (a + m);
      if (b >= l && b <= r) break;
      if (l > b) {
        n = 51;
        break;
      }
    }
    if (n >= 50) {
      cout << "-1" << '\n';
      continue;
    }
    cout << n + 1 << " " << a << " ";
    ll sum = a, tot = 0;
    for (int i = 1; i < n; ++i) {
      ll r = min((b - (1ll << (n - 1)) - (1ll << (n - 1)) * a - tot) /
                     (1ll << (n - i - 1)),
                 m - 1);
      cout << sum + r + 1 << " ";
      tot += r * (1ll << (n - 1 - i));
      sum += (sum + r + 1);
    }
    cout << b << '\n';
  }
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to the following 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vi = vector<int>;
using pii = pair<int, int>;
const int MAXN = 1e5 + 10;
template <typename T>
istream &operator>>(istream &is, vector<T> &v) {
  for (auto i = begin(v); i != end(v); i++) is >> *i;
  return is;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  for (auto i = begin(v); i != end(v); i++)
    os << *i << (i == end(v) - 1 ? "\n" : " ");
  return os;
}
signed main() {
  int t;
  cin >> t;
  while (t--) {
    ll a, b, m;
    cin >> a >> b >> m;
    if (a >= b) {
      if (a > b) cout << "-1" << '\n';
      if (a == b) cout << "1 " << a << '\n';
      continue;
    }
    int n;
    for (n = 1; n <= 50; ++n) {
      ll l = (1ll << (n - 1)) * (a + 1), r = (1ll << (n - 1)) * (a + m);
      if (b >= l && b <= r) break;
      if (l > b) {
        n = 51;
        break;
      }
    }
    if (n >= 50) {
      cout << "-1" << '\n';
      continue;
    }
    cout << n + 1 << " " << a << " ";
    ll sum = a, tot = 0;
    for (int i = 1; i < n; ++i) {
      ll r = min((b - (1ll << (n - 1)) - (1ll << (n - 1)) * a - tot) /
                     (1ll << (n - i - 1)),
                 m - 1);
      cout << sum + r + 1 << " ";
      tot += r * (1ll << (n - 1 - i));
      sum += (sum + r + 1);
    }
    cout << b << '\n';
  }
  return 0;
}

```