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 a, b, m;
long long s[110];
long long bit[110];
long long D = (long long)1e14;
bool ok(int n) {
  for (int i = 1; i <= n; ++i) {
    if (s[i] < 1 || s[i] > D) {
      return false;
    }
  }
  long long tot = s[1];
  for (int i = 2; i <= n; ++i) {
    if (s[i] - tot < 1 && s[i] - tot > m) {
      return false;
    }
  }
  return true;
}
void solve() {
  if (a == b) {
    printf("1 %lld\n", a);
    return;
  }
  for (int i = 2; i <= 50; ++i) {
    s[1] = a;
    long long tot = s[1];
    for (int j = 2; j <= i; ++j) {
      s[j] = tot + 1;
      tot += s[j];
    }
    if (s[i] > b) {
      continue;
    }
    long long remind = b - s[i];
    for (int j = 2; j <= i; ++j) {
      long long now = bit[i - j + 1];
      long long need = remind / now;
      need = min(need, m - 1);
      remind -= need * now;
      for (int k = j; k <= i; ++k) {
        s[k] += need * bit[k - j + 1];
      }
    }
    if (remind == 0 && ok(i)) {
      printf("%d", i);
      for (int j = 1; j <= i; ++j) {
        printf(" %lld", s[j]);
      }
      puts("");
      return;
    }
  }
  puts("-1");
}
int main() {
  bit[1] = 1;
  bit[2] = 1;
  for (int i = 3; i <= 55; ++i) {
    bit[i] = bit[i - 1] << 1;
  }
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%lld%lld%lld", &a, &b, &m);
    solve();
  }
  return 0;
}
