Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int nxt() {
  int x;
  cin >> x;
  return x;
}
long long int nxtl() {
  long long int x;
  cin >> x;
  return x;
}
void SieveOfEratosthenes(int n, vector<int> &primes) {
  bool prime[n + 1];
  memset(prime, true, sizeof(prime));
  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
  for (int p = 2; p <= n; p++)
    if (prime[p]) primes.push_back(p);
}
long long int max(long long int a, long long int b) {
  if (a > b) return a;
  return b;
}
long long int power(long long int x, long long int y, long long int mod) {
  long long int temp;
  if (y == 0) return 1;
  temp = power(x, y / 2, mod);
  if (y % 2 == 0)
    return (temp * temp) % mod;
  else
    return (((x * temp) % mod) * temp) % mod;
}
pair<long long int, long long int> cal(long long int x1, long long int y1,
                                       long long int x2, long long int y2);
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = nxt();
  while (t--) {
    long long int a = nxtl(), b = nxtl(), m = nxtl();
    vector<pair<long long int, long long int> > v;
    bool f = 0;
    long long int mul = 1;
    int xyz = 1;
    if (a == b) {
      cout << "1 " << a << endl;
      continue;
    }
    for (int i = 0; i < 50; i++) {
      long long int x = mul * a + mul;
      long long int y = mul * a + mul * m;
      v.push_back({x, y});
      if (b >= x && b <= y) {
        f = 1;
        break;
      }
      mul *= 2;
      xyz++;
      if (!xyz) break;
      if (mul * (a + m) > 4e18) xyz = -1;
    }
    if (!f) {
      cout << "-1\n";
      continue;
    }
    vector<long long int> prefix;
    auto it = cal(b - m, b - 1, v[v.size() - 1].first - 1,
                  v[v.size() - 1].second - m);
    prefix.push_back(it.first);
    v.pop_back();
    vector<long long int> ans;
    for (int i = v.size() - 1; i >= 0; i--) {
      long long int x = v[i].first - 1, y = v[i].second - m;
      long long int x2 = (prefix[prefix.size() - 1] - m + 1) / 2;
      long long int y2 = (prefix[prefix.size() - 1] - 1) / 2;
      auto it = cal(x, y, x2, y2);
      prefix.push_back(it.first);
    }
    prefix.push_back(0);
    reverse((prefix).begin(), (prefix).end());
    for (int i = 1; i < prefix.size(); i++)
      ans.push_back(prefix[i] - prefix[i - 1]);
    ans.push_back(b);
    cout << ans.size() << " ";
    for (auto i : ans) cout << i << " ";
    cout << endl;
  }
  return 0;
}
pair<long long int, long long int> cal(long long int x1, long long int y1,
                                       long long int x2, long long int y2) {
  pair<long long int, long long int> ret = {max(x1, x2), min(y1, y2)};
  return ret;
}