Fractional XOR Maximization

Define the bitwise XOR operation $\oplus$ for any two real numbers $x$ and $y$ as follows. $x \oplus y = \lim_{n \to \infty}{\frac{\lfloor2^nx\rfloor \oplus \lfloor 2^ny\rfloor}{2^n}}$ The $\oplus$ on the right side is the integer bitwise XOR. For example, $\frac{5}{8} \oplus \frac{3}{8} = \frac{3}{4}$, $\frac{1}{3} \oplus \frac{1}{7} = \frac{4}{9}$, and $\frac{1}{5} \oplus \frac{1}{7} = \frac{6}{65}$.

The supremum of a set $X \subseteq \mathbb{R}$, written $\sup X$, is the least real number that is greater than or equal to every element of $X$.

You are given four non-negative rational numbers $a$, $b$, $c$, and $d$. Find $\sup\{x \oplus y : x \in [a, b], y \in [c, d]\}$. In other words, find the least value that is greater than or equal to every XOR value formed by an element of $[a, b]$ and an element of $[c, d]$.

The first line contains $T$ ($1 \le T \le 100$), the number of test cases. Each test case has two lines. The first line contains four integers $a_{num}$, $a_{denom}$, $b_{num}$, $b_{denom}$ ($0 \le a_{num}, b_{num} \le 10^{17}$, $1 \le a_{denom}, b_{denom} \le 30$), which describe $a = \frac{a_{num}}{a_{denom}}$ and $b = \frac{b_{num}}{b_{denom}}$. The second line describes $c$ and $d$ in the same format. Every fraction is irreducible, and $a \le b$ and $c \le d$ hold.

Print $T$ lines. The $i$-th line contains the numerator and denominator of $x = \frac{x_{num}}{x_{denom}}$ for the $i$-th test case, separated by a single space, as an irreducible fraction.

In the first test case, the answer is $2$. Choosing $1$ from the first interval and $1 - 2^p$ from the second makes the XOR value arbitrarily close to $2$ for a large enough integer $p$, but no value equal to or greater than $2$ is reachable.