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;
const int N = 1e6 + 10, MOD = 1e9 + 7;
const long long oo = 1LL << 60;
long long mul(long long a, long long b) {
  if (a > oo / b) return oo;
  return a * b;
}
long long sum(long long a, long long b) {
  if (a + b > oo) return oo;
  return a + b;
}
bool test(long long a, long long b, long long m, int sz) {
  if (sz == 1) {
    if (a == b) {
      cout << 1 << " " << a << '\n';
      return true;
    }
    return false;
  }
  vector<long long> factors;
  for (int i = 0; i < sz - 1; i++) {
    factors.push_back((1LL << i));
  }
  vector<long long> rs(sz + 1);
  reverse(factors.begin(), factors.end());
  factors.push_back(1);
  b -= mul(a, factors[0]);
  for (int i = 1; i < factors.size(); i++) {
    if (b < 0) {
      return false;
    }
    long long lo = 1, hi = m;
    while (lo < hi) {
      long long mid = (lo + hi + 1) / 2;
      long long curr = mul(mid, factors[i]);
      for (int k = i + 1; k < factors.size(); k++) {
        curr = sum(factors[k], curr);
      }
      if (curr > b) {
        hi = mid - 1;
      } else {
        lo = mid;
      }
    }
    b -= mul(factors[i], lo);
    rs[i] = lo;
  }
  if (b != 0) {
    return false;
  }
  cout << sz << " " << a << " ";
  long long sum = a;
  for (int i = 1; i < factors.size(); i++) {
    cout << sum + rs[i] << " ";
    sum += sum + rs[i];
  }
  cout << '\n';
}
void solve() {
  long long a, b, m;
  cin >> a >> b >> m;
  bool flag = false;
  for (int sz = 1; sz <= 50; sz++) {
    bool ans = test(a, b, m, sz);
    if (ans) {
      flag = true;
      break;
    }
  }
  if (!flag) cout << -1 << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int tc;
  cin >> tc;
  while (tc--) {
    solve();
  }
}