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;
mt19937 rnd(time(0));
bool isPrime(long long a) {
  long long i;
  for (i = 2; i * i <= a; i++) {
    if (a % i == 0) {
      return 0;
    }
  }
  return (a != 1);
}
bool isPalindrom(const string &s, int l, int r) {
  for (int i = 0; i < (r - l + 1) / 2; i++) {
    if (s[l + i] != s[r - i]) {
      return 0;
    }
  }
  return 1;
}
template <class type1>
type1 gcd(type1 a, type1 b) {
  type1 t;
  while (b) {
    t = b;
    b = a % b;
    a = t;
  }
  return a;
}
long long positiveMod(long long v, long long md) {
  v = v % md;
  if (v < 0) {
    v += md;
  }
  return v;
}
void solve() {
  long long a, b, m, n, i, j, exSum, ta, dif, oneSum, k, tsum;
  cin >> a >> b >> m;
  if (a == b) {
    cout << 1 << " " << a << "\n";
    return;
  }
  if (b - a <= m) {
    cout << 2 << " " << a << " " << b << "\n";
    return;
  }
  for (i = 2; i <= 50; i++) {
    vector<long long> res(i + 1);
    exSum = a;
    ta = a;
    for (j = 1; j <= i; j++) {
      ta = exSum;
      ta++;
      exSum += ta;
      if (ta > b) {
        break;
      }
    }
    if (ta > b) {
      break;
    }
    exSum = a;
    ta = a;
    for (j = 1; j <= i; j++) {
      ta = exSum;
      ta += m;
      exSum += ta;
      if (ta >= b) {
        break;
      }
    }
    if (ta < b) {
      continue;
    }
    exSum = a;
    ta = a;
    for (j = 1; j <= i; j++) {
      ta = exSum;
      ta++;
      exSum += ta;
    }
    dif = b - ta;
    tsum = a;
    res[0] = a;
    for (j = 1; j <= i; j++) {
      if (i == j) {
        oneSum = 1;
      } else {
        oneSum = (1LL << (i - j - 1));
      }
      k = dif / oneSum;
      k = min(m - 1, k);
      dif -= k * oneSum;
      res[j] = tsum + k + 1;
      tsum += res[j];
    }
    cout << i + 1 << " ";
    for (auto it : res) {
      cout << it << " ";
    }
    cout << "\n";
    return;
  }
  cout << -1 << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t, tt;
  cin >> tt;
  for (t = 0; t < tt; t++) {
    solve();
  }
  return 0;
}
