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;
const long long mod = 1e9 + 7;
const int infi = 1147483600;
const long long infl = 1e18 + 5;
const char ENDL = '\n';
const long long MOD = 998244353;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int q;
  cin >> q;
  for (int xx = 0; xx < (int)q; xx++) {
    long long a, b, m;
    cin >> a >> b >> m;
    vector<long long> ans = {a};
    if (a == b) {
      cout << 1 << " " << a << ENDL;
      continue;
    }
    if (b - a >= 1 && b - a <= m) {
      cout << 2 << " " << a << " " << b << ENDL;
      continue;
    }
    int res = -1;
    for (int i = 0; i < (int)49; i++) {
      long long bk = b - (1ll << i) * (a + 1);
      if (bk < 0) break;
      if (0 <= bk && bk <= (1ll << i) * (m - 1)) {
        res = i;
        break;
      }
    }
    if (res == -1) {
      cout << -1 << ENDL;
      continue;
    }
    long long bk = b - (1ll << res) * (a + 1);
    vector<long long> r(res + 1, 1);
    cout << res + 2 << " " << a << " ";
    for (int i = 0; i < (int)res; i++) {
      long long ri = bk / (1ll << (res - 1 - i));
      ri = min(m - 1, ri);
      bk -= (1ll << (res - 1 - i)) * ri;
      r[i] += ri;
    }
    r[res] += bk;
    long long now = a;
    for (int i = 0; i < (int)res + 1; i++) {
      cout << now + r[i] << " ";
      now <<= 1;
      now += r[i];
    }
    cout << ENDL;
  }
}
