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;
typedef long long ll;
const bool DEBUGMRK = false;
int T;
ll a, b, m;
ll bin[70], lb, rb;
ll ans[70];
void solve(int x) {
  printf("%d", x);
  for (int i = 2; i <= x; i++) ans[i] = 1;
  ll tot = b - lb;
  (*bin) = 1;
  for (int i = x; i >= 3; i--) bin[++*bin] = (1LL << i - 3);
  bin[++*bin] = 1;
  for (int i = 2; i <= x; i++) {
    ll tmp = min(tot / bin[i], m - 1);
    ans[i] += tmp;
    tot -= bin[i] * tmp;
  }
  tot = a;
  printf(" %I64d", a);
  for (int i = 2; i <= x; i++) {
    printf(" %I64d", tot + ans[i]);
    tot = tot + tot + ans[i];
  }
  putchar('\n');
}
int main() {
  scanf("%d", &T);
  if (DEBUGMRK) printf("%d\n", T);
  while (T--) {
    scanf("%I64d%I64d%I64d", &a, &b, &m);
    if (DEBUGMRK) printf("%I64d %I64d %I64d\n", a, b, m);
    if (a == b) {
      printf("1 %I64d\n", a);
      continue;
    }
    bool flg = false;
    for (int i = 2; i <= 50; i++) {
      lb = (a + 1) * (1LL << i - 2), rb = (a + m) * (1LL << i - 2);
      if (lb > b) break;
      if (b > rb) continue;
      solve(i);
      flg = true;
      break;
    }
    if (!flg) puts("-1");
  }
  return 0;
}
