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 n, m, x, y, z;
long long q;
long long anss[100];
bool ansprint(long long a, long long b, long long koyghor, long long extra,
              long long m) {
  memset(anss, 0, sizeof(anss));
  for (long long i = 1; i < koyghor; i++) {
    anss[i] = 1;
  }
  long long tmp = koyghor - 1;
  tmp = (1LL << tmp) - 1;
  if (tmp >= extra) return false;
  long long hgst = 0;
  for (long long i = 0; i <= koyghor - 2; i++) {
    long long xxx = m * (1LL << i);
    hgst += xxx;
  }
  hgst += m;
  if (extra > hgst) return false;
  anss[koyghor] = extra - tmp;
  if (anss[koyghor] > m) {
    anss[koyghor - 1] += anss[koyghor] - m;
    anss[koyghor] = m;
  }
  for (long long i = koyghor - 1; i >= 2; i--) {
    if (anss[i] > m) {
      long long tmp = anss[i] - m;
      anss[i] = m;
      if (tmp % 2) tmp++, anss[i]--;
      anss[i - 1] += tmp / 2;
    }
  }
  for (long long i = 1; i <= koyghor; i++) {
    if (anss[i] == 0 || anss[i] > m) return false;
  }
  cout << koyghor + 1 << ' ' << a;
  long long ttt = 0, prv = 0;
  for (long long i = 1; i <= koyghor; i++) {
    prv = anss[i];
    long long xx = i - 1;
    xx = (1LL << xx);
    xx = xx * a;
    anss[i] += xx + ttt;
    ttt = ttt * 2 + prv;
    printf(" %lld", anss[i]);
  }
  printf("\n");
  return true;
}
int main() {
  cin >> q;
  while (q--) {
    long long a, b, m;
    scanf("%lld %lld %lld", &a, &b, &m);
    if (a == b) {
      cout << 1 << ' ' << a << endl;
      continue;
    }
    if (a + m >= b) {
      cout << 2 << ' ' << a << ' ' << b << endl;
      continue;
    }
    long long koyghor = -1, extra;
    for (long long i = 1, j = 1;; i = i * 2, j++) {
      long long tmp = a * i;
      if (tmp >= b) break;
      long long xx = b - tmp;
      bool ttt = ansprint(a, b, j, xx, m);
      if (ttt) {
        koyghor = j;
        extra = xx;
        break;
      }
    }
    if (koyghor == -1) {
      printf("-1\n");
    }
  }
  return 0;
}
