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 int INF = 0x3f3f3f3f;
template <class T>
void read(T &val) {
  T x = 0;
  T bz = 1;
  char c;
  for (c = getchar(); (c < '0' || c > '9') && c != '-'; c = getchar())
    ;
  if (c == '-') {
    bz = -1;
    c = getchar();
  }
  for (; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - 48;
  val = x * bz;
}
const int mod = 998244353;
const int maxn = 1e6 + 10;
long long x, y, m, t;
int check(vector<long long> a) {
  long long res = a[0] * x;
  long long se = 0;
  for (int i = 1; i < a.size(); i++) {
    se += a[i];
  }
  long long mi = res + se;
  long long mx = res + se * m;
  if (mx < y) return 0;
  if (mi > y) return -1;
  return 1;
}
long long ans[maxn];
void print(vector<long long> res) {
  long long nw = res[0] * x;
  long long sum = 0;
  for (int i = 1; i <= res.size() - 1; i++) {
    sum += res[i];
  }
  long long ned = y - nw;
  for (int i = 1; i <= res.size() - 1; i++) {
    sum -= res[i];
    ans[i] = min(m, (ned - sum) / res[i]);
    ned -= ans[i] * res[i];
  }
  long long lst = x;
  printf(" %lld", x);
  for (int i = 1; i <= res.size() - 1; i++) {
    nw = lst + ans[i];
    lst += nw;
    printf(" %lld", nw);
  }
  puts("");
}
int main() {
  read(t);
  while (t--) {
    read(x);
    read(y);
    read(m);
    if (x == y) {
      printf("1 %lld\n", x);
      continue;
    }
    vector<long long> a;
    a.push_back(1);
    a.push_back(1);
    int flag = 1;
    for (int i = 2; i <= 50; i++) {
      int res = check(a);
      if (res == -1) {
        flag = 0;
        break;
      } else if (res == 1) {
        printf("%d", i);
        print(a);
        break;
      }
      for (int j = 0; j < a.size() - 1; j++) {
        a[j] *= 2;
      }
      a.push_back(1);
    }
    if (!flag) puts("-1");
  }
  return 0;
}
