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;
void fun() {}
long long int md = 998244353;
long long int __gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return __gcd(b, a % b);
}
long long int poww(long long int a, long long int b, long long int md) {
  if (b < 0) return 0;
  if (a == 0) return 0;
  long long int res = 1;
  while (b) {
    if (b & 1) {
      res = (1ll * res * a) % md;
    }
    a = (1ll * a * a) % md;
    b >>= 1;
  }
  return res;
}
long long int poww(long long int a, long long int b) {
  if (b < 0) return 0;
  if (a == 0) return 0;
  long long int res = 1;
  while (b) {
    if (b & 1) {
      res = (1ll * res * a);
    }
    a = (1ll * a * a);
    b >>= 1;
  }
  return res;
}
void ainp(long long int arr[], long long int n) {
  for (long long int i = 1; i <= n; i++) cin >> arr[i];
}
long long int divide(long long int a, long long int b, long long int md) {
  long long int rr = a * (poww(b, md - 2, md));
  rr %= md;
  return rr;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  fun();
  long long int q;
  cin >> q;
  while (q--) {
    long long int a, b, m;
    cin >> a >> b >> m;
    if (a == b) {
      cout << "1 " << a << "\n";
      continue;
    }
    long long int lows[51], highs[51], ph[51], pl[51];
    highs[0] = lows[0] = ph[0] = pl[0] = a;
    long long int i = 0;
    long long int fg = 0;
    while (lows[i] < b) {
      lows[i + 1] = pl[i] + 1;
      highs[i + 1] = ph[i] + m;
      pl[i + 1] = pl[i] + lows[i + 1];
      ph[i + 1] = ph[i] + highs[i + 1];
      i++;
      if (lows[i] <= b && highs[i] >= b) {
        fg = 1;
        break;
      }
    }
    if (!fg) {
      cout << "-1\n";
      continue;
    }
    long long int ans[i + 1];
    ans[i] = b;
    long long int vv = 1ll << (i - 1);
    long long int lowsum = vv * a + vv - 1;
    ans[i - 1] = max(lowsum, b - m);
    for (long long int j = i - 2; j >= 0; j--) {
      long long int val = ceil((ans[j + 1] - m) / 2.0);
      vv = 1ll << (j);
      lowsum = vv * a + vv - 1;
      ans[j] = max(lowsum, val);
      ans[j + 1] -= ans[j];
    }
    cout << i + 1 << " ";
    long long int sum = 0;
    for (long long int j = 0; j <= i; j++) cout << ans[j] << " ";
    cout << "\n";
  }
  return 0;
}
