Reversed Engineered from Uniqueness (Cop's Thread) Given the output of the cop's program (o), the byte-count (n) and the number of unique bytes (c) used, come up with a corresponding piece of code that is n bytes long with c unique bytes which matches the cop's output o.

This is the cops thread. Post solutions that are to-be-cracked here.
The robbers thread is located here.

Cops should post solutions like this:
#[Language], `n` Bytes, `c` Unique Bytes (c*n points) [Cracked](Link-To-Crack)

    [Output]

---
    
(Optional spoiler)


Rules

You may not take any input for your program.
The program must use at least 1 byte, but cannot exceed 255 bytes.
The output itself is also limited to 255 bytes.
The program must have consistent output results when executed multiple times.
If your submission is not cracked within 7 days, you may mark it as &quot;safe&quot;.

When marking it safe, post the intended solution and score it as c*n.
Only mark an answer as &quot;safe&quot; if you really want the +15 for an accepted answer; it's more fun to see how long you can go.



Winning

The uncracked post with the lowest c*n score, wins the cop's thread.
This will be decided after 10 safe answers, or a few weeks.

Caveats

If you feel cocky, you may tell the user the algorithm using a spoiler tag.


Uncracked Submissions:


fetch("https://api.stackexchange.com/2.2/questions/147635/answers?order=desc&amp;sort=activity&amp;site=codegolf&amp;filter=!.Fjs-H6J36vlFcdkRGfButLhYEngU&amp;key=kAc8QIHB*IqJDUFcjEF1KA((&amp;pagesize=100").then(x=&gt;x.json()).then(data=&gt;{var res = data.items.filter(i=&gt;!/^#.*cracked/im.test(i.body_markdown)).map(x=&gt;{const matched = /^ ?#{1,3} ?(?:(?:(?:\[|&lt;a href ?= ?".*?"&gt;)([^\]]+)(?:\]|&lt;\/a&gt;)(?:[\(\[][a-z0-9/:\.]+[\]\)])?)|([^, ]+)).*[^\d](\d+) ?\[?(?:.*(?:byte|block|codel)s?)(?:\](?:\(.+\))?)? ?(?:\(?(?!no[nt][ -]competing)\)?)?/gim.exec(x.body_markdown);if(!matched){return;}return {link: x.link, lang: matched[1] || matched[2], owner: x.owner}}).filter(Boolean).reverse().forEach(ans=&gt;{var tr = document.createElement("tr");var add = (lang, link)=&gt;{var td = document.createElement("td");var a = document.createElement("a");a.innerHTML = lang;a.href = link;td.appendChild(a);tr.appendChild(td);};add(ans.lang, ans.link);add(ans.owner.display_name, ans.owner.link);document.querySelector("tbody").appendChild(tr);});});
&lt;html&gt;&lt;body&gt;&lt;h1&gt;Uncracked Submissions&lt;/h1&gt;&lt;table&gt;&lt;thead&gt;&lt;tr&gt;&lt;th&gt;Language&lt;/th&gt;&lt;th&gt;Author&lt;/th&gt;&lt;/tr&gt;&lt;/thead&gt;&lt;tbody&gt;&lt;/tbody&gt;&lt;/table&gt;&lt;/body&gt;&lt;/html&gt;
Haskell, 173 bytes, 8 unique, 1384 points, safe

&quot;[tag:reverse_engineering]&quot;

It looks like this answer wants to provide a relevant tag to the question, while using only 8 different bytes.
As usual, your solution should work by being pasted in the code block on TIO: Try it online!

Solution
'[':___':_'':_':':':__':__:__'':__:__':____:__:'_':__:___:_':_'_:___:__:__:__':_'_:___:_':']':[]
_:_:_'':_:_:_:__:_:_':_:_'_:_:_:_:_:___:_:_:_:__':____:___':_:__'':_=['_'..]

Yes, this is valid Haskell code: Try it online!
How it works
The underscore _ is used as a wild card in Haskell's pattern matching. ' is used for characters, e.g. 'a'. However, both _ and ' are also part of the allowed characters for identifier names (Within some restrictions, e.g. ' cannot occur at the beginning of the name.). Therefore __, _', _'_, _'' and so on are all valid identifier names. Using some more descriptive names, the above code becomes
'[':t:a:g:':':r:e:v:e:r:s:e:'_':e:n:g:i:n:e:e:r:i:n:g:']':[]
_:_:a:_:_:_:e:_:g:_:i:_:_:_:_:n:_:_:_:r:s:t:_:v:_=['_'..]

The first line yields the string &quot;[tag:reverse_engineering]&quot; if each variable is assigned to the correct character. This assignment is achieved in the second line: ['_'..] yields the string &quot;_`abcdefghijklmnopqrstuvwxyz{|}~ ... &quot;, witch is matched on the pattern _:_:a:_:_:_:e:_:g:_:i:_:_:_:_:n:_:_:_:r:s:t:_:v:_ to get the desired assignment a='a', e='e' and so on.