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;
long long a, b, m;
void solve() {
  cin >> a >> b >> m;
  if (a == b) {
    cout << 1 << ' ' << a << '\n';
    return;
  }
  if (m == 1) {
    vector<long long> ans;
    long long val = a;
    long long sm = a;
    ans.push_back(a);
    while (true) {
      val = sm + 1;
      sm += val;
      ans.push_back(val);
      if (val > b) {
        cout << -1 << '\n';
        return;
      }
      if (val == b) {
        break;
      }
    }
    cout << (int)ans.size() << ' ';
    for (auto e : ans) {
      cout << e << ' ';
    }
    cout << '\n';
    return;
  }
  if (b - a <= m) {
    cout << 2 << ' ' << a << ' ' << b << '\n';
    return;
  }
  for (int len = 3; len <= 50; len++) {
    long long mn = a, sm = a;
    bool pos = true;
    for (int rep = 0; rep < len - 1; rep++) {
      mn = sm + 1;
      if (mn > b) {
        pos = false;
        break;
      }
      sm += mn;
    }
    if (!pos) {
      continue;
    }
    vector<long long> ans(len, 0);
    long long diff = b - mn;
    long long val = 0;
    for (int i = 49; i > 0; i--) {
      if ((1LL << i) & diff) {
        val += (1LL << i);
      }
      if (i > len - 3) {
        continue;
      }
      if (val / (1LL << i) > m - 1) {
        val -= (m - 1) * (1LL << i);
        diff -= (1LL << i) * (m - 1);
        ans[len - 2 - i] += (m - 1);
      } else {
        ans[len - 2 - i] += (val / (1LL << i));
        val = 0;
        diff -= val;
      }
    }
    val += (diff & 1);
    if (val > m - 1) {
      val -= m - 1;
      diff -= m - 1;
      ans[len - 1] += m - 1;
    } else {
      ans[len - 1] += val;
      diff -= val;
      val = 0;
    }
    if (val > m - 1) {
      val -= m - 1;
      diff -= m - 1;
      ans[len - 2] += m - 1;
    } else {
      ans[len - 2] += val;
      diff -= val;
      val = 0;
    }
    if (val == 0) {
      sm = a, val = a;
      cout << len << ' ' << val << ' ';
      for (int rep = 0; rep < len - 1; rep++) {
        val = sm + ans[rep + 1] + 1;
        sm += val;
        cout << val << ' ';
      }
      cout << '\n';
      return;
    }
  }
  cout << -1 << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
