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;
const long long M = 1e5 + 10, M2 = 1e6 + 10, oo = 1e9 + 7;
const long long MC = (1 << 20) + 5;
void precomp() {}
void solve(int64_t T, int64_t tc) {
  int64_t a, b, m;
  cin >> a >> b >> m;
  if (a == b) {
    cout << 1 << " " << a << endl;
    return;
  }
  vector<int64_t> v, ad(55, 0), have(55, 0);
  int64_t k;
  v.push_back(a);
  for (k = 0; k <= 52; k++) {
    if ((1LL << k) * (a + 1) > b) break;
    v.push_back((1LL << k) * (a + 1));
  }
  m--;
  k--;
  int64_t sub = b - v.back();
  int64_t need = 0;
  for (int64_t i = 52; i >= 0; i--) {
    need *= 2;
    if ((sub >> i) & 1LL) {
      need++;
    }
    if (i + 1 <= k) {
      int64_t c = min(m, need);
      ad[i + 1] += c;
      need -= c;
    }
  }
  if (need <= m) {
    ad[0] = need;
    need = 0;
  }
  if (need) {
    cout << -1 << endl;
    return;
  }
  assert(v.size() == k + 2);
  cout << v.size() << " " << v[0] << " ";
  for (int i = k; i >= 0; i--) {
    int64_t sm = 0;
    for (int j = k; j > i; j--) sm += ad[j];
    ad[i] += sm;
  }
  for (int i = 1; i <= k + 1; i++) cout << v[i] + ad[k - i + 1] << " ";
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  precomp();
  int64_t T;
  cin >> T;
  for (int64_t tc = 1; tc <= T; tc++) {
    solve(T, tc);
  }
  return 0;
}
