Reversing Key Algorithm
You forgot to mention that $x$ is in fact $64$ bits long. Write $x$ in binary notation:
$$ x_{63} x_{62} \ldots x_1 x_0. $$
We have
$$
\begin{align*}
op_0 &amp;= x_{63} \cdots x_{56} \\
op_1 &amp;= x_{55} \cdots x_{48} \\
op_2 &amp;= x_{19} \cdots x_{16} \\
op_3 &amp;= x_{23} \cdots x_{20} \\
op_4 &amp;= x_{27} \cdots x_{24} \\
op_5 &amp;= x_{31} \cdots x_{28} \\
op_6 &amp;= x_{35} \cdots x_{32} \\
op_7 &amp;= x_{39} \cdots x_{36} \\
op_8 &amp;= x_{43} \cdots x_{40} \\
op_9 &amp;= x_{47} \cdots x_{44} \\
op_{10} &amp;= x_{15} \cdots x_8
\end{align*}
$$
All these, in binary notation.

I'll let you take it from here.