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;
using ll = long long;
int n;
ll a, b, B, m, r[55];
bool ok(int i, ll v) {
  ll S = B;
  if (i == n) {
    S -= r[i];
    S += v;
    return S <= b;
  }
  S -= r[i] << (n - i - 1);
  S += v << (n - i - 1);
  return S <= b;
}
ll bs(ll lo, ll hi, int i) {
  if (hi - lo <= 1) {
    for (ll j = hi; j >= lo; --j)
      if (ok(i, j)) return j;
    return lo - 1;
  }
  ll m = lo + hi >> 1;
  if (ok(i, m)) return bs(m, hi, i);
  return bs(lo, m, i);
}
int main() {
  int tt;
  scanf("%d", &tt);
  while (tt--) {
    scanf("%lld%lld%lld", &a, &b, &m);
    if (a == b) {
      printf("1 %lld\n", a);
      continue;
    }
    bool flag = false;
    for (int k = 2; k <= 50; ++k) {
      n = k - 1;
      for (int i = 1; i <= n; ++i) r[i] = 1;
      bool flag1 = true;
      ll sum = a;
      for (int i = 1; i <= n && flag1; ++i) {
        ll cur = sum + r[i];
        if (cur > b) flag1 = false;
        B = cur;
        sum += cur;
      }
      if (!flag1 || B > b) continue;
      for (int i = 1; i <= n; ++i) {
        ll bes = bs(1, m, i);
        if (i == n) {
          B -= r[i];
          B += bes;
        } else {
          B -= r[i] << (n - i - 1);
          B += bes << (n - i - 1);
        }
        r[i] = bes;
      }
      sum = a;
      vector<ll> vans{a};
      for (int i = 1; i <= n; ++i) {
        ll cur = sum + r[i];
        vans.push_back(cur);
        sum += cur;
      }
      if (vans.back() == b) {
        flag = true;
        printf("%d", (int)vans.size());
        for (ll x : vans) printf(" %lld", x);
        printf("\n");
        break;
      }
    }
    if (!flag) printf("-1\n");
  }
  return 0;
}
