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;
using pii = pair<int, int>;
using lint = long long;
const int MOD = 1e9 + 7, INF = 987654321;
const lint LINF = 987654321987654321;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
const int mxn = 55;
int tc, cnt;
int n;
lint pw2[mxn];
lint ans[mxn];
lint r[mxn];
lint a, b, m;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  pw2[0] = 1;
  for (int i = 1; i <= 50; i++) pw2[i] = pw2[i - 1] * 2LL;
  cin >> tc;
  while (tc--) {
    cin >> a >> b >> m;
    if (a == b) {
      cout << 1 << ' ' << a << '\n';
      continue;
    }
    lint k;
    for (k = 2; k <= 51; k++) {
      if (b < pw2[k - 2] * a + pw2[k - 2]) {
        k = 51;
        break;
      }
      if (((pw2[k - 2] * a + pw2[k - 2]) <= b) &&
          (b <= (pw2[k - 2] * a + pw2[k - 2] * m))) {
        break;
      }
    }
    if (k > 50) {
      cout << "-1\n";
      continue;
    }
    if (k == 2) {
      cout << k << ' ' << a << ' ' << b << '\n';
      continue;
    }
    memset(ans, 0, sizeof(ans));
    memset(r, 0, sizeof(r));
    ans[1] = a;
    for (int i = 2; i <= k; i++) {
      r[i] = 1;
      ans[i] = a * pw2[i - 2];
    }
    lint diff = b - ans[k] - pw2[k - 2];
    for (int i = 2; i <= k - 1; i++) {
      lint cur = min(diff / pw2[k - 1 - i], m - 1);
      r[i] += cur;
      diff -= cur * pw2[k - 1 - i];
    }
    if (diff) r[k] += diff;
    for (int i = 2; i <= k; i++) {
      ans[i] += r[i];
      for (int j = 2; j < i; j++) {
        ans[i] += r[j] * pw2[i - 1 - j];
      }
    }
    cout << k;
    for (int i = 1; i <= k; i++) cout << ' ' << ans[i];
    cout << '\n';
  }
  return 0;
}
