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 ca[100];
long long cr[100][100], r[100], ans[100], sca[100], scr[100][100], sum[100];
int main() {
  int q;
  cin >> q;
  while (q--) {
    memset(ca, 0, sizeof ca);
    memset(cr, 0, sizeof cr);
    memset(r, 0, sizeof r);
    memset(ans, 0, sizeof ans);
    memset(sca, 0, sizeof sca);
    memset(scr, 0, sizeof scr);
    memset(sum, 0, sizeof sum);
    long long a, b, m;
    cin >> a >> b >> m;
    if (a == b) {
      cout << 1 << " " << a << endl;
      continue;
    }
    ca[1] = sca[1] = 1;
    int flag = 0;
    for (long long i = 2; i <= 50; i++) {
      ca[i] = sca[i - 1];
      sca[i] = sca[i - 1] + ca[i];
      long long sumr = 0;
      for (long long j = 2; j <= i; j++) {
        if (j == i) {
          cr[j][i] = scr[j][i] = 1;
        } else {
          cr[j][i] = scr[j][i - 1];
          scr[j][i] = scr[j][i - 1] + cr[j][i];
        }
        sumr += cr[j][i];
      }
      if (ca[i] >= b || sumr >= b) break;
      long long minx = ca[i] * a + sumr;
      long long maxx = ca[i] * a + sumr * m;
      if ((double)ca[i] <= (double)(b - sumr) / (double)a && b <= maxx) {
        long long res = b - minx;
        for (int j = 2; j <= i; j++) {
          r[j] = min(m - 1, res / cr[j][i]);
          res -= r[j] * cr[j][i];
          if (res == 0) break;
        }
        if (res != 0) break;
        ans[1] = sum[1] = a;
        for (int j = 2; j <= i; j++) {
          for (int k = 1; k <= j; k++)
            ans[j] = sum[j - 1] + cr[k][j] * (r[k] + 1);
          sum[j] = sum[j - 1] + ans[j];
        }
        cout << i << endl;
        for (int j = 1; j <= i; j++) {
          cout << ans[j] << " ";
        }
        cout << endl;
        flag = 1;
        break;
      }
    }
    if (!flag) cout << -1 << endl;
  }
}
