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;
const int NMAX = 51;
const long long MAX_LONG_LONG = 1LL << 62;
int Q;
long long a, b, m;
long long pow2[NMAX];
long long safeAdd(long long a, long long b) {
  if (a >= MAX_LONG_LONG - b) {
    return MAX_LONG_LONG;
  }
  return a + b;
}
long long safeMul(long long a, long long b) {
  if (a >= MAX_LONG_LONG / b) {
    return MAX_LONG_LONG;
  }
  return a * b;
}
long long fMin(int N, long long a) {
  return safeAdd(safeMul(pow2[N - 2], a), pow2[N - 2]);
}
long long fMax(int N, long long a) {
  return safeAdd(safeMul(pow2[N - 2], a), safeMul(pow2[N - 2], m));
}
bool solutionExists(int N, long long a, long long b) {
  return fMin(N, a) <= b && b <= fMax(N, a);
}
void recons(int N, long long a, long long b, vector<long long>& sol) {
  assert(N >= 2);
  assert(solutionExists(N, a, b));
  if (N <= 2) {
    sol.push_back(a);
    sol.push_back(b);
    return;
  }
  long long l = 1, r = m + 1;
  while (r - l > 1) {
    long long mid = l + (r - l) / 2;
    if (safeAdd(fMin(N - 1, a + mid), safeMul(pow2[N - 3], a)) > b) {
      r = mid;
    } else {
      l = mid;
    }
  }
  long long maxLeft = l;
  l = maxLeft - 1;
  r = m;
  while (r - l > 1) {
    long long mid = l + (r - l) / 2;
    if (safeAdd(fMax(N - 1, a + mid), safeMul(pow2[N - 3], a)) < b) {
      l = mid;
    } else {
      r = mid;
    }
  }
  long long minRight = r;
  assert(maxLeft <= minRight);
  sol.push_back(a);
  recons(N - 1, a + maxLeft, b - a * pow2[N - 3], sol);
}
int main() {
  pow2[0] = 1;
  for (int i = 1; i < NMAX; i++) {
    pow2[i] = pow2[i - 1] * 2;
  }
  cin >> Q;
  for (int query = 0; query < Q; query++) {
    cin >> a >> b >> m;
    if (a == b) {
      cout << "1 " << b << "\n";
      continue;
    }
    bool foundSol = false;
    vector<long long> sol;
    for (int N = 2; N <= 50 && !foundSol; N++) {
      if (solutionExists(N, a, b)) {
        cout << N;
        recons(N, a, b, sol);
        long long carry = 0;
        for (size_t i = 0; i < sol.size(); i++) {
          carry = carry * 2 + (i >= 2 ? sol[i - 2] : 0);
          cout << " " << sol[i] + carry;
        }
        cout << "\n";
        foundSol = true;
      }
    }
    if (!foundSol) {
      cout << "-1\n";
    }
  }
  return 0;
}
