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;
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
void shoot() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
long long power(long long n, long long pwr) {
  long long ans = 1LL;
  while (pwr) {
    if (pwr & 1) ans = (ans * n);
    n = (n * n);
    pwr >>= 1;
  }
  return ans;
}
int main() {
  shoot();
  long long q;
  cin >> q;
  while (q--) {
    long long i, j, k, a, b, c, m, sum, pre, req, prev;
    cin >> a >> b >> m;
    if (a == b) {
      cout << 1 << " " << a << endl;
      continue;
    }
    if (b <= a + m) {
      cout << 2 << " " << a << " " << b << endl;
      continue;
    }
    for (i = 3; i <= 50; i++) {
      c = b - a * (1LL << i - 2);
      if ((1LL << i - 2) > c) {
        i = 51;
        break;
      }
      if ((1LL << i - 2) <= c && c <= (1LL << i - 2) * m) {
        long long r[i - 1];
        for (j = 0; j < i - 1; j++) {
          r[j] = 1;
        }
        req = c - (1LL << i - 2);
        if (req == 0) {
          sum = a, pre = a;
          cout << i << " " << pre << " ";
          for (j = 0; j < i - 1; j++) {
            pre += r[j];
            cout << pre << " ";
            sum += pre;
            pre = sum;
          }
          cout << endl;
          break;
        }
        j = i - 3;
        k = 0;
        while (j >= 0 && k < i - 2 && req != 0) {
          r[k] += min(req / (1LL << j), m - 1);
          req -= min(req / (1LL << j), m - 1) * (1LL << j);
          j--;
          k++;
        }
        r[i - 2] += min(req, m - 1);
        req -= min(req, m - 1);
        if (req != 0) {
          cout << -1 << endl;
          break;
        }
        sum = a, pre = a;
        cout << i << " " << pre << " ";
        for (j = 0; j < i - 1; j++) {
          pre += r[j];
          cout << pre << " ";
          sum += pre;
          pre = sum;
        }
        cout << endl;
        break;
      }
    }
    if (i == 51) cout << -1 << endl;
  }
  return 0;
}
