Explain this PGM encryption as simply as possible pls$!$PGM example $A_5$ of order $60$

Encryption of $49$:

$49=40+5+4$

${}^{-1}=(4,1,2)$

$={}_{}(4,1,2)=$

$(1)(2)(354)*(1)(23)(45)*(15432)=$

$(154)(2)(3)($i$.$e$.,$ hat$())$

Let $(154)(2)(3)={}_{3p_3}^{}*{}_{2p_2}*{}_{1p_1}$

Then ${}_{1p_1}=([1,5,3,4,2])$ so

$p_1=4$ and $p_{14}{m}_1=4.$

${}_{3p_3}*{}_{2p_2}=(154)(2)(3)*{}_{14}^{-1}=$

$(154)(2)(3)*(12435)=(1)(24)(35).$

Then ${}_{2p_2}=(1)(243)(5)$ so

$p_2=2$ and $p_{22}{m}_2=10.$

${}_{3p_3}=(1)(24)(35)*{}_{22}^{-1}=$

$(1)(24)(35)*(1)(234)(5)=$

$(1)(2)(354),$ so

$p_3=1$ and $p_{31}{m}_3=20.$

So $20+10+4=34.$

i$.$e$.,E_{,}(49)=34.$

Note:

To figuring out ${}_{1p_1},$ look for $1?,$ which should be unique,

for ${}_{2p2},$ look for $2?($unique$),$ since $11$ must be already. $...$

for ${}_{i{p}_i},$ look for $i?($unique$),$ since $11,22,$dots,$(i-1)(i-1)$ must be already.