$1.$ The following cipher text was generated using a simple substitution algorithm:

Xli uymgo fvsar jsb nyqtw sziv xli pepd hsk$.$ Mr xli wlehsav sj xli sprimklx, e qcwxivmsyw jmkyvi aexglih jvsq xli hmwxergi. Gpyiw alvi wgexxivih evsyrh xli wgiri, fyx rsri qehi wirwi ex jmvwx kpergi. Xli hixigxmzi orsa xlex wspzmrk xImw gewi asyph viuymvi qsvi xler nywx vsyxmri mrziwxmkexmsr. Imklx tistpi leh wmir wsqixlmrk xlex rmklx$,$ fyx xlimv wxsvmiw hmhr$\text{'}$x uymxi qexgl yt$.$ E vih wgevj, e fvsoir aexgl, erh e qcwxivmsyw rsxi aivi xli srpc tmigiv sj izmhirgi. Xmqi aew vyrrmrk syx$,$ erh xli tviwwyvi xs wspzi xli gewi aew qsyrxmrk$.$ Ew xli wyr vswi sr ersxliv hec, xli hixigxmzi viepmDih xlex wsqixmqiw xli sfzmsyw erwaiv mw Imhhir mr tpemr wmklx$.$

Decrypt this message. Hints:

$1.$ As you know, the most frequently occurring letter in English is $\backslash (\backslash$mathbf$\{$e$\}\backslash ).$ Therefore, the first or second $[$or perhaps third?$)$ most common character in the message is likely to stand for e$.$ Also, e is often seen in pairs $[$e$.$g$.,$ meet, fleet, speed, seen, been, agree, etc.$).$ Try to find a character in the cipher text that decodes to $\backslash (\backslash$mathbf$\{$e$\}\backslash ).$

$2.$ The most common word in English is "the." Use this fact to guess the characters that stand for $\backslash (\backslash$mathbf$\{$t$\}\backslash )$ and $\backslash (\backslash$mathbf$\{$h$\}\backslash ).$

$3.$ Decipher the rest of the message by deducing additional words.

Note: The resulting message is in English but may not make much sense on a first reading.

Show the step$-$by$-$step process of getting the result.