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 N = 51;
long long a, b, m;
long long ans[N];
void add(long long i, long long x, long long &b, long long len) {
  ans[i] += x;
  if (i == len - 1) b -= x;
  for (long long j = i + 1; j < len; ++j) {
    ans[j] += x;
    if (j == len - 1) b -= x;
    x <<= 1;
  }
}
void print(long long len) {
  if (len == 1) {
    cout << "1 " << a << '\n';
    return;
  }
  for (long long i = 0; i < N; ++i) ans[i] = 0;
  add(0, a, b, len);
  for (long long i = 1; i < len; ++i) add(i, 1, b, len);
  for (long long i = 1; i < len; ++i) {
    long long k = 1;
    if (i < len - 1) k <<= len - 2 - i;
    add(i, min(m - 1, b / k), b, len);
  }
  cout << len << ' ';
  for (long long i = 0; i < len; ++i) cout << ans[i] << ' ';
  cout << '\n';
}
long long l[N], r[N];
void solve() {
  cin >> a >> b >> m;
  l[0] = r[0] = a;
  long long lsum = a, rsum = a;
  for (long long i = 1; i < N; ++i) {
    l[i] = lsum + 1;
    r[i] = rsum + m;
    lsum += l[i];
    rsum += r[i];
  }
  for (long long i = 0; i < N; ++i) {
    if (l[i] > b) {
      cout << "-1\n";
      return;
    }
    if (l[i] <= b && b <= r[i]) {
      print(i + 1);
      return;
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long q;
  cin >> q;
  while (q--) {
    solve();
  }
}
