The Collatz sequence is defined as: $a_{i+1}=\{\genfrac{}{}{0ex}{}{\frac{a_i}{2}}{3a_i+1}\genfrac{}{}{0ex}{}{\text{if }{a}_i\text{ is even}}{\text{if }{a}_i\text{ is odd}}$.

The Collatz conjecture states that starting from any positive integer, the sequence eventually reaches the cycle $1,4,2,1,\dots$.
We shall define the sequence prefix $p(n)$ for the Collatz sequence starting with $a_1=n$ as the sub-sequence of all numbers not a power of $2$ ($2^0=1$ is considered a power of $2$ for this problem). For example:
$p(13)=\{13,40,20,10,5\}$
$p(8)=\{\}$
Any number invalidating the conjecture would have an infinite length sequence prefix.

Let $S_m$ be the set of all sequence prefixes of length $m$. Two sequences $\{a_1,a_2,\dots ,a_m\}$ and $\{b_1,b_2,\dots ,b_m\}$ in $S_m$ are said to belong to the same prefix family if $a_i<a_j$ if and only if $b_i<b_j$ for all $1\le i,j\le m$.

For example, in $S_4$, $\{6,3,10,5\}$ is in the same family as $\{454,227,682,341\}$, but not $\{113,340,170,85\}$.
Let $f(m)$ be the number of distinct prefix families in $S_m$.
You are given $f(5)=5$, $f(10)=55$, $f(20)=6771$.

Find $f(90)$.