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 a, b, m;
long long check(long long sum, long long val, long long mov) {
  long long lo = val, hi = val, los = sum + val, his = sum + val;
  mov--;
  while (mov--) {
    lo = los + 1;
    los += lo;
    hi = his + m;
    his += hi;
  }
  if (hi < b) {
    return 1;
  } else if (lo > b) {
    return -1;
  } else {
    return 0;
  }
}
long long Bsrch(long long sum, long long pval, long long mov) {
  long long lo = sum + 1, hi = sum + m;
  while (hi >= lo) {
    long long mid = (hi + lo) / 2;
    long long va = check(sum, mid, mov);
    if (va == 1) {
      lo = mid + 1;
    } else if (va == -1) {
      hi = mid - 1;
    } else {
      return mid;
    }
  }
  return -1;
}
long long checker() {
  long long mo = 1;
  long long lo = a, hi = a, los = a, his = a;
  while (hi < b) {
    lo = los + 1;
    los += lo;
    hi = his + m;
    his += hi;
    mo++;
  }
  if (lo > b)
    return -1;
  else
    return mo;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long q;
  cin >> q;
  while (q--) {
    cin >> a >> b >> m;
    if (a == b) {
      cout << "1 " << a << endl;
      continue;
    }
    if (a + m >= b) {
      cout << "2 " << a << " " << b << endl;
      continue;
    }
    long long mov = checker();
    if (mov == -1) {
      cout << "-1" << endl;
      continue;
    }
    vector<long long> ans;
    mov--;
    long long sum = a;
    ans.push_back(a);
    while (1) {
      long long val = Bsrch(sum, ans[ans.size() - 1], mov);
      mov--;
      sum += val;
      ans.push_back(val);
      if (mov == 0) break;
    }
    cout << ans.size() << " ";
    for (long long i : ans) cout << i << " ";
    cout << endl;
  }
  return 0;
}
