Vulcan: a literary dis-allusion Issue Two

IiEFg

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PPPM=PA=L<=PAP

=P==PL<PP<PM= 

PA=Q<M=\AA\QPAP

L< 

P<L=PP<=<

AAP@=Qu\M@@=tpPA@P AP<<PQuFPQpPAPLPp=RQ uL@@p 

PP\P=\QP=QL<

P<P 

\<A=RL@P<

L@A@P==PA= APAA 

Pp=P<PQL<===<<PP@A\

=QP@QuS<sFp@QpP=A\ 

PAPP===L<P<L<\

P@Qu@p 

\PP<L=\P<P=<=L@P

@QuS<<QpP<PP L\pP<P 

\AAPAALL

\A<L<P<PP@ MPP<@=P<==<=P PM@<P==P PPM=AA<

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\<PPM@Qu\MPP p 

PP=P=<AAP

MPL@PPPPP<=@ \@QuPPAPQp P=P\=QA=\PA =PA==\ P@QuP=P@p 

PPPAQPQ

P<QP=P=P

Bc