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;
long long f[55][55];
long long res[55];
bool fill(int x, int k) {
  if (x == 0) return false;
  if (res[x - 1] == 0) {
    fill(x - 1, k);
    if (res[x - 1] == 0) return false;
    res[x - 1]--;
    if (res[x - 1] == 0) fill(x - 1, k);
    if (res[x - 1] == 0) return false;
    if (f[k][x - 1] == f[k][x])
      res[x]++;
    else
      res[x] += 2;
    return true;
  } else {
    res[x - 1]--;
    if (res[x - 1] == 0) fill(x - 1, k);
    if (f[k][x - 1] == f[k][x])
      res[x]++;
    else
      res[x] += 2;
    if (res[x - 1] == 0) return false;
    return true;
  }
}
bool check(long long x, long long m, int k, long long a) {
  if (x < 0) return false;
  memset(res, 0, sizeof(res));
  for (int i = 2; i <= k; i++) {
    if (f[k][i] > 0) {
      res[i] = x / f[k][i];
      res[i] = min(res[i], m);
      x -= f[k][i] * res[i];
    }
  }
  if (x > 0) return false;
  for (int i = 2; i <= k; i++) {
    if (res[i] == 0) {
      fill(i, k);
    }
  }
  for (int i = 2; i <= k; i++)
    if (res[i] < 1 || res[i] > m) return false;
  cout << k << " ";
  long long ans[55];
  memset(ans, 0, sizeof(ans));
  ans[1] = a;
  for (int i = 2; i <= k; i++) {
    ans[i] = res[i];
    for (int j = 1; j < i; j++) {
      ans[i] += ans[j];
    }
  }
  for (int i = 1; i <= k; i++) cout << ans[i] << " ";
  cout << endl;
  return true;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int q = 0;
  f[1][1] = 1;
  for (int i = 2; i < 50; i++) {
    f[i][i] = 1;
    for (int j = 0; j < i; j++) {
      for (int k = 0; k < 50; k++) {
        f[i][k] += f[j][k];
      }
    }
  }
  cin >> q;
  while (q--) {
    long long a, b, m;
    cin >> a >> b >> m;
    bool ok = false;
    for (int i = 1; i < 50; i++) {
      if (b / a < f[i][1]) break;
      if (check(b - a * f[i][1], m, i, a)) {
        ok = true;
        break;
      }
    }
    if (!ok) cout << -1 << endl;
  }
  return 0;
}
