Minimal maximum byte for Turing Completeness Summary
We already have a challenge for the minimal number of characters for Turing completeness. But what about the minimal maximum byte?
Challenge
For a language of your choice, find a subset of characters that allows your language to be Turing complete with the minimal maximum byte value.
Example
If your language is using the UTF8 code page and Turing complete using only the characters abcd(), these characters have the byte values 97, 98, 99, 100, 40, 41, and the maximum there is d = 100, so your score would be 100.
Scoring
Your score is the maximum byte value of your Turing complete subset, expressed in the code page of your language's interpreter. For most languages this will be either LATIN-1 or UTF8, but if your language uses a different code page (such as Jelly) use that to score instead.
In the unlikely event that you require multibyte characters to be Turing complete, just use the maximum byte in the multibyte sequence to score. For example, if your UTF8 language required Ȧ, this is encoded as the literal byte sequence 200, 166, so it would score 200.
The lowest score wins.
Notes

The minimum score is 0, the maximum is 255.
For some languages (cough) this challenge will be trivial to score low on. I urge you to upvote interesting answers over trivial ones.
Provide an explanation of why your subset is Turing complete.
Your subset only needs to be Turing complete, it does not need to be able to use every feature of the language.
Haskell, score 61 (=)
Characters used: !#$%&amp;()=
The SKI combinator calculus can be implemented in Haskell with nothing but basic function definition, using !#%&amp; as identifiers. Infix function application $ is used to save on parentheses and strip down one character from both S and fix. Finally, K only takes two arguments and can be more shortly defined as an infix operator.
S combinator: s x y z = x z (y z)
(!)(#)(%)(&amp;)=(#)(&amp;)$(%)(&amp;)

K combinator: k x y = x
(!)#($)=(!)

fix combinator: fix f = f (fix f)
(&amp;)(!)=(!)$(&amp;)(!)

Since Haskell is a strongly typed language, the fixed-point combinator fix is needed in order to make (typed) combinatory logic Turing-complete.
The I combinator is not strictly required since it is extensionally equivalent to SKK, but it can be defined as (*)(!)=(!).
Try it online!