That set is a harmful set

While studying set theory, Seunghyun ran into the following family of sets.

$C_0 = [0, 1]$

$C_n = \left\{\, a + \sum_{i=1}^{n} \frac{a_i}{3^i} \;\middle|\; 0 \le a \le \frac{1}{3^n},\; a_i \in \{0, 2\} \,\right\} \quad (n \in \mathbb{N})$

$C = \bigcap_{n=0}^{\infty} C_n$

$C$ made no sense to him, so he planned two jobs. The first job picks a rational number and decides whether it belongs to $C$. If it does not, the second job finds the set with the smallest subscript $n$ that fails to contain the number.

After doing this by hand a few times he got tired of it and decided to use a computer. Then he got tired of the coding too, so he handed the work to you, his junior. Stuck with the job, you decided to write an incomplete checker to play a trick on him.

If the answers look too strange he will not fall for it, so the output has to stay plausible. The program checks membership only for $C_0$ through $C_{10}$, and if all eleven sets contain the number it declares the number to be in $C$. Write that incomplete program.

The first line contains the number of test cases $T$.

Each of the next $T$ lines contains the numerator $a$ and the denominator $b$ of a rational number, separated by a space. ($0 \le a \le 100000$, $1 \le b \le 100000$)

Print one line per test case.

If some set among $C_0$ through $C_{10}$ does not contain $\frac{a}{b}$, print the smallest such subscript $n$. If all eleven sets contain it, print -1.

$C_1 = \left[0, \frac{1}{3}\right] \cup \left[\frac{2}{3}, 1\right]$, and $C_2 = \left[0, \frac{1}{9}\right] \cup \left[\frac{2}{9}, \frac{1}{3}\right] \cup \left[\frac{2}{3}, \frac{7}{9}\right] \cup \left[\frac{8}{9}, 1\right]$.

$\frac{1}{2}$ belongs to $C_0$ but not to $C_1$. $\frac{1}{6}$ belongs through $C_1$ and drops out at $C_2$.

A number with numerator 0 lies in every set: take $a = 0$ and every $a_i = 0$ in the definition.

If $\frac{a}{b}$ is greater than 1, it already fails at $C_0$, so the answer is 0.