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>
#pragma GCC omptimize("unroll-loops")
#pragma optimize("SEX_ON_THE_BEACH")
using namespace std;
using ll = long long int;
using ull = unsigned long long int;
using dd = long double;
using ldd = long double;
using si = short int;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
ll get_seed(string s) {
  ll ans = 0;
  for (int i = 0; i < s.size(); ++i) ans += s[i];
  return ans;
}
void solve() {
  0;
  ;
  ll a, b, m;
  cin >> a >> b >> m;
  vector<ll> arr = {a};
  while (arr.back() < b) {
    ll s = 0;
    for (ll i : arr) s += i;
    arr.push_back(s + 1);
  }
  if (arr.back() == b) {
    cout << arr.size() << " ";
    for (ll i : arr) cout << i << " ";
    cout << '\n';
    return;
  }
  arr.pop_back();
  for (int i = 1; i < arr.size(); ++i) {
    0;
    ;
    ll need = arr.back() - b;
    need *= -1;
    0;
    ;
    vector<ll> a(arr.size());
    a[i] = 1;
    for (int j = i + 1; j < arr.size(); ++j) {
      for (int k = 0; k < j; ++k) {
        a[j] += a[k];
      }
    }
    0;
    ;
    need /= a.back();
    need = min(need, m - 1);
    0;
    ;
    a[i] = need;
    for (int j = i + 1; j < arr.size(); ++j) {
      a[j] = 0;
      for (int k = 0; k < j; ++k) a[j] += a[k];
    }
    for (int i = 0; i < arr.size(); ++i) arr[i] += a[i];
  }
  if (arr.back() == b) {
    cout << arr.size() << " ";
    for (ll i : arr) cout << i << " ";
  } else {
    cout << -1;
  }
  cout << '\n';
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  int t;
  cin >> t;
  for (int i = 0; i < t; ++i) solve();
  return 0;
}
