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 long long N = 1e3 + 10;
long long sum[N];
long long vis[N];
int main() {
  long long q, a, b, k, m, c;
  cin >> q;
  while (q--) {
    memset(vis, 0, sizeof vis);
    cin >> a >> b >> m;
    if (a == b) {
      cout << 1 << ' ';
      cout << a << endl;
      continue;
    }
    sum[1] = a;
    long long ss = a;
    for (int i = 2; ss + 1 <= b; i++) {
      sum[i] = ss + 1;
      ss += sum[i];
      k = b - sum[i];
      c = i - 2;
    }
    for (int i = c - 1; i >= 0; i--) {
      long long d = k / (1ll << i);
      if (d <= (m - 1)) {
        vis[c + 1 - i] = d;
        k -= ((1ll << i) * d);
      } else {
        vis[c + 1 - i] = m - 1;
        k -= ((1ll << i) * (m - 1));
      }
    }
    for (int i = 0; i >= 0; i--) {
      long long d = k / (1ll << i);
      if (d <= (m - 1)) {
        vis[c + 2] = d;
        k -= ((1 << i) * d);
      } else {
        vis[c + 2] = m - 1;
        k -= ((1ll << i) * (m - 1));
      }
    }
    if (k)
      cout << -1 << endl;
    else {
      cout << c + 2 << ' ';
      long long tt = 0;
      for (int i = 1; i <= c + 2; i++) {
        vis[i] += tt;
        tt += vis[i];
        sum[i] += vis[i];
        cout << sum[i] << ' ';
      }
      cout << endl;
    }
  }
}
