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;
template <typename _T>
inline void read(_T &f) {
  f = 0;
  _T fu = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') fu = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    f = (f << 3) + (f << 1) + (c & 15);
    c = getchar();
  }
  f *= fu;
}
template <typename T>
void print(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x < 10)
    putchar(x + 48);
  else
    print(x / 10), putchar(x % 10 + 48);
}
template <typename T>
void print(T x, char t) {
  print(x);
  putchar(t);
}
template <typename T>
struct hash_map_t {
  vector<T> v, val, nxt;
  vector<int> head;
  int mod, tot, lastv, last_id;
  T lastans;
  bool have_ans;
  hash_map_t(int md = 0) {
    head.clear();
    v.clear();
    val.clear();
    nxt.clear();
    tot = 0;
    mod = md;
    nxt.resize(1);
    v.resize(1);
    val.resize(1);
    head.resize(mod);
    have_ans = 0;
  }
  void clear() {
    for (register int i = 1; i <= tot; i++) head[v[i] % mod] = 0;
    tot = 0;
    v.clear();
    val.clear();
    nxt.clear();
    v.resize(1);
    val.resize(1);
    nxt.resize(1);
    have_ans = 0;
  }
  bool count(int x) {
    int u = x % mod;
    for (register int i = head[u]; i; i = nxt[i]) {
      if (v[i] == x) {
        have_ans = 1;
        lastv = x;
        last_id = i;
        lastans = val[i];
        return 1;
      }
    }
    return 0;
  }
  void ins(int x, int y) {
    if (count(x)) {
      val[last_id] = y;
      lastans = y;
      return;
    }
    int u = x % mod;
    nxt.push_back(head[u]);
    head[u] = ++tot;
    v.push_back(x);
    val.push_back(y);
  }
  int qry(int x) {
    if (have_ans && lastv == x) return lastans;
    count(x);
    return lastans;
  }
};
long long ans[55];
long long a, b, m;
int T;
int main() {
  read(T);
  while (T--) {
    read(a);
    read(b);
    read(m);
    if (a == b) {
      print(1, ' '), print(a, '\n');
      continue;
    }
    long long now = 1;
    bool ok = 0;
    for (register int len = 2; len <= 50; len++) {
      long long qaq = b - a * now - now;
      if (qaq < 0) break;
      long long now2 = now;
      for (register int i = 2; i < len; i++) {
        now2 >>= 1;
        long long tmp = min(m - 1, qaq / now2);
        qaq -= now2 * tmp;
        ans[i] = tmp;
      }
      if (qaq >= m) {
        now <<= 1;
        continue;
      }
      ans[len] = qaq;
      ok = 1;
      long long sum = a;
      print(len, ' ');
      print(a, ' ');
      for (register int i = 2; i <= len; i++) {
        ++ans[i];
        ans[i] += sum;
        sum += ans[i];
        print(ans[i], ' ');
      }
      putchar('\n');
      break;
    }
    if (!ok) print(-1, '\n');
  }
  return 0;
}
