Peel away the layers - Robbers This is the Robbers thread. For the Cops challenge, see here
As the robbers, your job is to crack the cops' answers. Choose an uncracked, unsafe cop answer and attempt to figure out which languages, in which order, the cop used.
Note that a crack is valid even if the languages are different. So long as the the languages you find, in the order you state them, produce the specified final phrase, the crack is valid, even if the languages are not the same as the intended or original cop answer.
If you are unable to fully crack a cop's answer and decide to abandon your attempt, you can (but are by no means expected to) comment underneath the cop's answer with the languages you've already figured out, in order to help other users.
Having fully cracked a cop's answer, post an answer to this thread listing all the languages, along with their outputs in the following format:
# [&lt;cop&gt;'s answer](&lt;link to answer&gt;), &lt;N&gt; layers

Languages used:
- Layer 1: [&lt;language 1&gt;](&lt;link to interpreter&gt;), outputted &lt;output&gt;
- Layer 2: [&lt;language 2&gt;](&lt;link to interpreter&gt;), outputted &lt;output&gt;
...
- Layer &lt;N&gt;: [&lt;language N&gt;](&lt;link to interpreter&gt;), outputted &lt;final phrase&gt;

---

&lt;anything else&gt;

After posting an answer, leave a comment below the cop's post to let them know it's been cracked. If you have \$&lt;50\$ reputation at the time of posting, and therefore cannot comment, I (or another able user) will be happy to do so for you, simply make a note of this in your answer and someone will take care of it.
This challenge can result in some very difficult answers to crack - more so than the standard cops-and-robbers - due to the fact that the number of languages a cop could've used increases exponentially per layer they have. Therefore, be sure to check out the Robbers' chatroom to get help cracking an answer, or to help others.
Your score is calculated the following way:

If you crack \$n\$ answers, with \$x_1, x_2, ..., x_n\$ languages being used (i.e. the first has \$x_1\$ languages, the second \$x_2\$ etc.), your score is

$$\frac{1}{n}\sum^n_{i=1}x_i$$
i.e. the total number of languages you cracked, divided by the total number of answers you cracked. The lower your score, the better. Don't bother editing your score into your answers, a winner will be declared when all cop answers are either cracked or safe.
Good luck!
Sisyphus's Answer, 33 layers
i=496;a=     0
print('gREETINGS!'.swapcase()if a==447775else'i=%3s;a=%6s%s'%(i-16,a+(int(1/2*2)*int('syd0s6ncczl7ka1ioletm9t1atoqjq2awpshwfxcorai94tcmzim91j2zzq2c08wxjmfmzr0x9euhjojhltjpay0g8t1o78y83a',36)&gt;&gt;i&amp;0xFFFF),open(__file__).read().strip()[14:]))

All layers are Python, in this order: 3, 3, 3, 3, 3, 2, 3, 3, 2, 2, 2, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2/3 (the last layer does not matter).
Special thanks to @Riolku for helping me find the right algorithm for subset sum (because I'm bad at computer science and Google)
Since there are so many layers, writing them all as a list would be rather difficult and take me a lot of time (and anyone who wants to verify it a ton of time), so instead you can verify it by running all 33 layers with this Bash program that compiles all of the layers together.
Let me know if this posting format is invalid.