\[ \let\oldlor\lor \renewcommand{\lor}{\; \oldlor \;} \let\oldland\land \renewcommand{\land}{\; \oldland \;} \renewcommand{\set}[1]{\{\, #1 \,\}} \renewcommand{\given}{\,\mid\,} \renewcommand{\abs}[1]{\lvert #1 \rvert} \renewcommand{\divs}{\!\;\mid\;\!} \renewcommand{\ndivs}{\!\;\nmid\;\!} \renewcommand{\betw}[3][1]{#1 \leq #2 \leq #3} \renewcommand{\mod}[1]{\ (\mathrm{mod}\ #1)} \renewcommand{\floor}[1]{\left \lfloor #1 \right \rfloor} \renewcommand{\ceil}[1]{\left \lceil #1 \right \rceil} \renewcommand{\t}[1]{\texttt{#1}} \renewcommand{\fori}[2][i]{\text{for } #1 = 0, 1, \dots, #2} \renewcommand{\x}[1]{\text{#1}} \renewcommand\concat{\mathbin{+\mkern-10mu+}} \DeclareMathOperator*{\CONCAT}{\concat} \DeclareMathOperator*{\SCC}{\|} \]
CodeForces 977D - Divide by three, multiply by two
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Model the problem as a graph, drawing directed edges from $i \longrightarrow j$, such that $i / 3 = j$ ($i$ must be a multiple of $3$) or $2i = j$, for each ordered pair $i, j$ where $i \not = j$.

Note that the problem gives the bound $1 \leq a_i \leq 10^{18}$, for $i = 0, \dots, n-1$, so we need to use long long. This gives room for integers up to $2^{63}-1$, which leaves us plenty of room (even if we're careless doing our checks; since, at worst, $3 * 10^{18} < 2^{63}-1$)

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977B Two-gram
977D Divide by three, multiply by two
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