#include <bits/stdc++.h>

#define int long long
using namespace std;

int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);

  int t; cin >> t;
  while (t--) {
    int n; cin >> n;
    vector<int> ans;
    for (int i = 1; (1LL << i) <= 2 * n; i++) {
      int j = i + 1, i2 = 1LL << i, j2 = 1LL << j;
      if ((n + i2) % (j2 - 1) == 0) { // (n + i2) / (j2 - 1) == x
        ans.push_back(((n + i2) / (j2 - 1)) * (j2 - 2) - i2 + 1);
      }
    }
    if (ans.empty()) cout << -1;
    for (auto& i : ans) cout << i << ' ';
    cout << '\n';
  }

  return 0;
}

I don't have a proof for that but it works like this: If number $$$N$$$ can be written as $$$(2^{k + 1} - 1)x - 2^k$$$ Then this number is in the answer: $$$(2^{k + 1} - 2)n - 2^k + 1$$$. Also $$$x$$$ and $$$k$$$ are positive integers. If you can come up with a proof, I would like to hear it.

for _ in range(int(input())):
    n=z=int(input())
    s,v,x = [2],0,1
    while(n):
        v+=x*((n&1)-(~n&1))
        s.append(v if v>0 else 4*x+v)
        x<<=1
        n>>=1
    c,m,y=1,2,1
    for i in s:
        if (m*c-i)%(m-1)==0 and m-2:print((m*c-i)//(m-1),end=" ");y=0
        c+=(z-i)//m+1
        m<<=1
    print(-1 if y else "")

I will update here after he explain it to me but as far as I know it is something like that: There will be approximately $$$logn$$$ tours around the cycle. In one cycle only 1 number can match with its index. In the first cycle numbers increase by 2, in the second by 4 and so on. But indices always increase by 1. If we know the first number chosen in the cycle we can find if there is a matching number. There is a bitwise trick to find that first number in the cycle but as I said I will update this part after his explanation.