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 ll = long long;
using i2 = pair<int, int>;
using ll2 = pair<ll, ll>;
ll R[59], S[59];
ll linear_pow(ll a, ll n) {
  ll res = 1;
  while (n--) res *= a;
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int q;
  cin >> q;
  while (q--) {
    ll a, b, m;
    cin >> a >> b >> m;
    if (a == b)
      cout << "1 " << a << "\n";
    else {
      R[1] = a - 1;
      bool ok = false;
      for (int k = 2; k <= 50; k++) {
        if (b < linear_pow(2, k - 2) * (a + 1))
          break;
        else {
          ll x = b - linear_pow(2, k - 2) * (a + 1);
          for (int l = 2; l < k; l++) {
            ll c = x / linear_pow(2, k - l - 1);
            R[l] = min(c, m - 1);
            x -= R[l] * linear_pow(2, k - l - 1);
          }
          if (x >= m) continue;
          R[k] = x;
          cout << k;
          for (int i = 1; i <= k; i++) {
            x = S[i - 1] + R[i] + 1;
            cout << " " << x;
            S[i] = S[i - 1] + x;
          }
          cout << "\n";
          ok = true;
          break;
        }
      }
      if (!ok) cout << "-1\n";
    }
  }
  return 0;
}
