Inverse Divisor Sums

Your friend Odd Even is obsessed with number theory. Every time he learns a new operation on numbers he spends hours applying it. Last year he learned Euler's totient function $\phi(n)$, which counts the positive integers up to $n$ that are relatively prime to $n$, and then computed $\phi(n)$ by hand for every integer $n$ from 1 to one million.

Recently he learned that the sum of all divisors of a number $N$ is given by the following formula.

$\text{sum of divisors}(N) = \prod_{i=1}^{r} \frac{p_i^{a_i+1} - 1}{p_i - 1}$

Here $p_1^{a_1}, p_2^{a_2}, \dots, p_r^{a_r}$ is the factorization of $N$ into prime factors, each $p_i$ is different, and $a_i$ is the largest exponent such that $p_i^{a_i}$ divides $N$.

Odd Even wants to run the function backwards. Given a positive integer $N$ he wants every positive integer $M$ whose sum of divisors is $N$, written out in increasing order. Doing that by hand would take far too long, so you decided to write a program for him instead.

Given a positive integer $N$, print every integer $M$ whose sum of divisors is $N$ in increasing order, or report that no such number exists.

The first line contains the number of test cases $T$. Each of the next $T$ lines contains one integer $N$.

• $1 \le T \le 100$
• $1 \le N \le 10^9$

For each test case, print every number whose sum of divisors is $N$ on one line in increasing order, with a single space between consecutive numbers. If no such number exists, print none! without the quotes.

The output can get very long, so collect it and write it out at once instead of printing line by line.