Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  char ch = getchar();
  long long s = 0, w = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    s = s * 10 + ch - '0';
    ch = getchar();
  }
  return s * w;
}
long long q;
long long s1[3010], s2[3010], S1[3010], S2[3010];
long long a, b, m;
long long T;
long long L[3010], R[3010];
inline long long check(long long val, long long now) {
  S1[now] = S2[now] = val;
  for (register long long i = now + 1; i <= T; i++) S1[i] = S1[i - 1] * 2 + 1;
  for (register long long i = now + 1; i <= T; i++) S2[i] = S2[i - 1] * 2 + m;
  if (S1[T] - S1[T - 1] <= b && S2[T] - S2[T - 1] >= b)
    return 1;
  else if (S2[T] - S2[T - 1] < b)
    return -1;
  else
    return 0;
}
inline void query(long long x) {
  long long l = L[x - 1] * 2 + 1, r = R[x - 1] * 2 + m;
  while (l <= r) {
    long long mid = (l + r) >> 1;
    int p = check(mid, x);
    if (p == 1)
      L[x] = mid, r = mid - 1;
    else if (p == -1)
      l = mid + 1;
    else
      r = mid - 1;
  }
  l = L[x - 1] * 2 + 1, r = R[x - 1] * 2 + m;
  while (l <= r) {
    long long mid = (l + r) >> 1;
    int p = check(mid, x);
    if (p == 1)
      R[x] = mid, l = mid + 1;
    else if (p == -1)
      l = mid + 1;
    else
      r = mid - 1;
  }
}
int main() {
  q = read();
  while (q--) {
    a = read(), b = read(), m = read();
    if (a == b) {
      printf("1 %I64d\n", a);
      continue;
    }
    s1[1] = s2[1] = a;
    long long flag = 0;
    for (register long long i = 2; i <= 50 && s1[i - 1] <= b; i++)
      s1[i] = s1[i - 1] * 2 + 1;
    for (register long long i = 2; i <= 50 && s1[i - 1] <= b; i++)
      s2[i] = s2[i - 1] * 2 + m;
    for (register long long i = 1; i <= 50 && s1[i - 1] <= b; i++) {
      if (s1[i] - s1[i - 1] <= b && s2[i] - s2[i - 1] >= b) {
        flag = i;
        break;
      }
    }
    if (!flag) {
      printf("-1\n");
      continue;
    }
    printf("%I64d ", flag);
    T = flag;
    L[1] = R[1] = a;
    for (register long long i = 2; i < flag; i++) query(i);
    for (register long long i = 1; i < flag; i++)
      printf("%I64d ", R[i] - R[i - 1]);
    printf("%I64d\n", b);
  }
  return 0;
}