The Communism Hireachy

1. ☭^(☭^(1/0)) = Capitalist Communism = NEUTRAL
2. ☭^(1/0) = CORRUPT ☭ = A structure with money = Capitalism
3. ☭^0=1=ERRAR (what I mean by rebounding point is when it will also be electrons, because it is loops forever.)
4. ☭^(2^(-2w-5)) = Electrons/Rebounding Point
5. ☭^(2^(-w-5)) = Electrons/Rebounding point (the next rebounding point is☭^(2^(-2w-5)))
6. ☭^(2^(-w-4)) = Electricity
7. ☭^(2^(-w-3)) = Calcalator
8. ☭^(2^(-w-2)) = God's Tool
9. ☭^(2^(-w-1)) = Lawbreaker
10. ☭^(2^-w)=Planks
11. ☭^(2^-20)=Strings
12. ☭^(2^-19)=Neutrino
13. ☭^(2^-18) = Leptons
14. ☭^(2^-17) = Protons, neutrons, and electrons
15. ☭^(2^-16) = Atoms
16. ☭^(2^-15) = Molecules
17. ☭^(1/16384) = Single=celled organisms
18. ☭^(1/8192) = Multi-celled orgnanisms
19. ☭^(1/4096) = Simple organisms
20. ☭^(1/2048) = Complex life
21. ☭^(1/1024) = Brain
22. ☭^(1/512) = Intelligence
23. ☭^(1/256) = Communication
24. ☭^(1/128) = Orginization
25. ☭^(1/64) = 3 Tribes
26. ☭^(1/32) = European Countries
27. ☭^(1/16) = Root of Evil = Between Capitalism and Communism=Socialism
28. ☭^(1/8) = Root of Eurasian Countries = Evil
29. ☭^(1/4) = Root of Communism = Eurasian Countries
30. ☭^(1/2) = Communism
31. ☭ = C(0) = USSR
32. ☭^1.5 = USSSSR (Union of Soviet Socialist Soviet Socialist Republics)
33. ☭^2  = USUCCER
34. ☭^w = GOOD
35. ☭^w^2 = FOREVER OP AND GOOD
36. ☭^w^3 = SUCC OP GOOD
37. ☭^w^4 = SUCC OP GOOD GOD
38. ☭^w^^2 = GOOOD
39. ☭^w^^^2 = GOOOOOOOOOOOOOOOOOOOOO...D
40. ☭^w{w}w = TRUE GOOD
41. ☭^FOOT(w) = C \* BIGGEST FOOT
42. ☭^☭ = THE FINAL GOOD
43. ☭^(☭+1) = THE REBORN GOOD
44. ☭^(2☭) = THE FINAL REBORN GOOD
45. ☭^(☭^2) = THE FINAL SUPERREBORN GOOD
46. ☭^^3 = THE FINAL MEGAREBIRTH GOOD
47. ☭{☭}☭ = THE VSR GOOD
48. {☭,☭,☭,☭} = THE USSR GOOD
49. {☭,☭,☭,☭,☭} = THE FINAL MERGED GOODEST NOT BEST JOHNY
50. ☭&☭ = Communist Planet
51. (X^(1/2048))[☭]&☭=C\*BIGGEST EXTREME ORGANISM=BIG C
52. (X^☭)[☭]&☭=C\*BIGGEST USSR=BIGGEST C
53. X^2[☭] &☭=C\*BIGGEST C\*BIGGEST EXTREME ORGANISM=BIGGEST C^2
54. X^3[☭] &☭=BIGGEST C^3
55. X^X[☭] &☭=BIGGEST C^C
56. X^^X[☭] &☭=BIGGEST C^^C
57. ☭&☭&☭=BIGGEST BIGGEST C
58. {☭,☭/2}=BIGGESTER C
59. {☭<>☭/☭<>☭}=BIGGESTEST C (A<>B=A&A&...A&A WITH B A'S)
60. {☭<>☭/☭<>☭/2}=BIGGESTESTER C
61. {☭<>☭/☭<>/☭<>☭}=BIGGESTESTESTEST C
62. {☭<>☭/☭<>☭/☭<>☭/2}=BIGGESTESTESTESTER C
63. {☭<>☭/☭<>☭/☭<>☭/☭<>☭}=BIGGESTESTESTESTEST C
64. ☭?=BIGG C (n!m={n<>n/n<>n/...n<>n} with m n<>n's)
65. ☭??=BIGGG C
66. FOOT(☭)=BIGGEST C'S FOOT EVER
67. FOOT(☭?)=BIGGEREST C'S FOOT EVER
68. ☭{1} = CCROUNTONILLION where ☭{n} repeats all steps starting from USSR to the step before the current step to ☭{n-1}. ☭{0} = ☭
69. ☭{2} = C(4)
70. ☭{4} = Kim Jong Un
71. ☭{16} = North Korea
72. ☭{256} = Reborn USSR
73. The next step follows the same pattern. We repeat over and over, until
74. ☭{☭} = USSR True Reborn
75. The next step follows the same pattern. We repeat over and over, until (we keep doing this)
76. ☭{☭{☭}} = USSR Third Reborn
77. ☭{} fixed point = USSR Superreborn
78. ☭{}phi(1,0,0... n times)phi(1,0) (see Durvenson's phi extension) = it's time to stop
79. f\_[☭](it's time to stop) = no more
80. f\_[[☭]](no more) where f\_a(b) only does ordinal rules if a is in [], else, it will be treated as a finite integer. Note that when we decompose a into a-1, unless a is not a limit ordinal, which in that case, we subsitute it, then a layer of [] will go. = Elemental 3
81. f\_[[[☭]]](Elemental 3) = Elememental 3
82. f\_[[[[☭]]]](Elemental 3) = Elemememental 3
83. Elemental 3(☭) = Elemlgntal (note that Elemental 3(0) =  f\_☭(it's time to stop), and  Elemental 3(n) = f\_[☭]... n layers(Elemental 3(n-1))
84. Elemental 3() fixed point = NOSCOPE
85. Elemental 4(n)=Elemental 3(NOSCOPE+n)
86. Elemental m+1 (n)=Elemental m(Elemental m(...Elemental m(NOSCOPE)...)) with n m+1's
87. a=Elemental a(a) fixed point=Chompster(1).
88. The communism scale is a notation now!
89. Chompster(n+1)=Elemental (Elemental n(Chompster(n))) (Elemental (Chompster(n))(n))
90. Elemental can be Carerelic[1] and Chompster can be Carerelic[2]
91. Carerelic[a](b)=Carerelic[a]^b (☭)
92. Carelic[NOSCOPE] (C(4))=C(5)
93. C(6)=Carelic[C(5)] (NOSCOPE)
94. C(7)=C(6) but replace $ with ☭
95. C(8)={C(7),C(7)/(1)2}
96. C(9)=C(8)\*BIGGEST EXTREME ORGANISM
97. C(10)=C(C(9))=C\*C(9)
98. C(1)=C(10)+0.999999... repeating with C(9) 9's
99. We continue defining C's
100. C(1048577/1048576)=C(1048576/1048576)^Planks
101. C(1048577.7/1048576)=C(1048577/1048576) but replace $ with (X^(C(1048576/1048576)/Planks))[☭]&☭
102. C(1048578/1048576)=v(C(1048577.7/1048576)) where we define the v function
	* v(1)=Gatorade=Swamp+AID=Swamp+SO MUCH C=SNAP IN THE CRINGE FRACTAL
	* v(2)=GridEverything=GridESNAP IN THE CRINGE FRACTALerything=Grid Exponential SNAP IN THE CRINGE FRACTAL spooky thing
	* v(3)=Number of s in the lyrics Spooky Scary Skeletons, and then take that number and then let s be the number of s. Then, v(3)=☭{☭{...{v(2)}...}} with s ☭{}
	* v(4)=v(3)+v(3)\*v(3)^v(3)^^v(3)^^^v(3)<insert v(3) arrows>tritritri...tri with v(3) tri using the tritri... in Fan-made Googology wiki https://fanmade-googology.wikia.com/wiki/Tritritri
	* v(5)=v(4) but replace all normal numbers n if less than w with The Big Gun https://fanmade-googology.wikia.com/wiki/Template:Salad\_Numbers\_Series
	* v(6)=Illuminati(v(5)) where illuminati is defined:
		+ Start with BOX\_Q https://fanmade-googology.wikia.com/wiki/BOX\_Q~
		+ BOX\_Q is the 5th number in the sequence, or BOX\_5.
		+ BOX\_M is BOX\_1, BOX\_O is BOX\_2, and so on
		+ Then, take BOX(BOX(...BOX(BOX\_OMEGA)...)) with n BOX functions where n is the number that is imputed in the illuminati() function
	* v(8)=Red Bull Advertisement in units of [plank seconds/v(7)]
	* v(8+n+1)=v(8+n)+C(8)+v(8)+v(n)+n+8+illuminati(n)+illuminati(8)
103. C(1048579/1048576)=-e^C(1048578/1048576)
104. C(1048580/1048576)=Proton(C(1048578/1048576)-C(1048579/1048576)) where Proton(n) is
	* Proton(0) = Elemental 3(19).
	* First, solve for the nth rebounding point. Call this Proton(n)[0].
	* Proton(n)[m] = The process for the step after m steps after the nth rebounding point, but replace all numbers with Proton(n)[m-1].
	* Repeat until you get to step 98. You have now solved it.
105. C(1048581/1048576)=Neutron(C(1048580/1048576),C(1048580/1048576)) where Neutron(n,m) is defined as
	* Start with Proton(n)&Proton(m).
	* Do that Proton^m(n) times, but with each step, you replace Proton(n) and Proton(m) with Proton(previous step).
106. C(1048582/1048576)=Electron(C(1048581/1048576),C(1048581/1048576))
	* **Electron(n,1) is the same as Proton(n,n) but replace the nth Rebounding Point with the BOX\_BOX\_BOX\_BOX\_BOX(BOX\_BOX(BOX\_BOX^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^... w ^s w(BOX(Proton(n,n)th Rebounding Point.**
	* **Electron(n,m) is Electron(n,1) but replace Proton(n,m) with Electron(n,n-1).**
	* **Electron(@a,b,1) = Electron(@,a,BOX\_BOX(Electron(29035890258902385903285903285903289032850328509238590238b))).**
	* **Electron(@,m) is Electron(n,1) but replace Proton(n,m) with Electron(@,m-1).**
107. C(1048583/1048576)=2^C(1048582/1048576)+P(C(1048581/1048576))+☭{C(1048580/1048576)} where P is power set
108. C(1048584/1048576)=**Electron(previous,previous...) with previous previouses.**
109. C(1048592/1048576)=**F\_[Previous](Previous)**
110. C(1+1/32768)=E(F\_[Previous](Previous))Previous#(F\_[Previous](Previous)) using ordinal Hyper E.
111. C(1+1/16384)=Dab(E(F\_[Previous](Previous))Previous#(F\_[Previous](Previous)),-E(F\_[Previous](Previous))Previous#(F\_[Previous](Previous))) where Dab(l+1,r-1)=Dab(l,r)+2 and Dab(0,0)=Neutron(C(1048580/1048576),C(1048580/1048576))
112. C(1+1/8192)=(Dab(E(F\_[Previous](Previous))Previous#(F\_[Previous](Previous)),-E(F\_[Previous](Previous))Previous#(F\_[Previous](Previous))))+round(C^-1(Dab(E(F\_[Previous](Previous))Previous#(F\_[Previous](Previous)),-E(F\_[Previous](Previous))Previous#(F\_[Previous](Previous)))))
113. C(1+1/4096)=**B**((Dab(E(F\_[Previous](Previous))Previous#(F\_[Previous](Previous)),-E(F\_[Previous](Previous))Previous#(F\_[Previous](Previous))))+round(C^-1(Dab(E(F\_[Previous](Previous))Previous#(F\_[Previous](Previous)),-E(F\_[Previous](Previous))Previous#(F\_[Previous](Previous)))))) where B(n) replace all ☭ with B, where B is defined as this:
	* Y=Z^3
	* X=Y\*Z+C(1048580/1048576)
	* W=X^^Y+{X,Z,X}
	* V=ceil(W/2)+ceil(X/2)+Y+1+Z
	* U=ME+ALL+U+V+W+X+Y+Z where ALL is Sbiis "**The Gathering**" and ME=U, making U a fixed point.
	* T=U\*V+Electron(W,X,Y,Z)
	* S=T+1+Proton(U^^V)^^^{W,X,Y,Z}
	* R=STW+BattleRoyale(UV)\*X\*Y\*Z where Battle Royale is defined like this:
		+ BattleRoyale(n)=e^(n\*Number of Fortnite Accounts created by the year n)
	* Q=BOX(BOX(BOX(BOX(BOX(BOX(BOX(BOX(BOX(R)+S)+T)+U)+V)+W)+X)+Y)+Z)
	* P=N\*P where N=☭{{Y,Z,((Q\*R)^(S\*T))^^((U\*V)^(W\*X))}} (N isn't actual N)
	* O=Smallest natural number n where 0\*n=0\*P\*Q\*R\*S\*T\*U\*V\*W\*X\*Y\*Z
	* N=N\*P\*Q\*R\*S\*T\*U\*V\*W\*X\*Y\*Z+O using a fixed point of N
	* M=McDonald(N)+N/2+O/4+P/8+Q/16+R/32+S/64+T/128+U/256+V/512+W/1024+X/2048+Y/4096+Z/8192 where McDonald have a scale or Menu. Here is the menu!
		+ 0=Wendy roasts!
		+ 1=Meat
		+ 2=Chicken
		+ 3=Fries
		+ w=Fruits
		+ w+1=Vegetables
		+ w2=Burger
		+ w^2=Whopper
		+ m=Big Mac (m=w^2/w)
		+ m2=Big Big Mac
		+ mc=Big Ma where c=C(1+1/65536)
		+ mc^2=Get out of Ma
		+ Energy=(mc)^2=(m^2)(c^2)=Shrek
		+ Electron=Energy^Energy=Sanic
		+ Sanic^Sanic=Sanic Hierarchy 2. Then, we make a SANIC hierarchy
			- SANIC(0)=Energy
			- SANIC(1)=SANIC(0)^SANIC(0)=SANIC
			- SANIC(n+1)=SANIC(n)^SANIC(n)
		+ McDonald(n)={(SANIC^n (n)),(SANIC^n (n)),43)} using BEAF burger!
	* L=Leg^M(FOOT^N(Leg^O(FOOT^P(Leg^Q(FOOT^R(Leg^S(FOOT^T(Leg^U(FOOT^V(Leg^W(FOOT^X(Leg^Y(FOOT^Z(BIGGEST FOOT EVER)))))))))))))) where BIGGEST FOOT EVER=C\*BIGGEST EXTREME ORGANISM and Leg(n)=FOOT^n (n)
	* K=(L)Kill((M)Kill((N)Kill((O)Kill((P)Kill((Q)Kill((R)Kill((S)Kill((T)Kill((U)Kill((V)Kill((W)Kill((X)Kill((Y)Kill(Z)))))))))))))) where (a)Kill(b)=a+b+1
	* J=PigLatin(KLMNOPQRSTUVWXYZ) where PigLatin(n) replace the last 2 letters with A\*Y where A is defined as this, just for now:
		+ Start with K
		+ Multiply by 1000
		+ Add Graham's number
		+ Apply the G() function or Graham's function
		+ Add Nathan Ho, or the number of Googology Wiki users by the year 10^100
		+ Then, apply Foot(Leg())
		+ Then, raise it to the 69th to get A
	* I=(-JKLMNOPQRSTUVWXYZ)^2
	* H=I+(-JKLMNOPQRSTUVWXYZ)^2
	* G=Graham^H({3,Graham^J({3,Graham^L({3,Graham^N({3,Graham^P({3,Graham^R({3,Graham^T({3,Graham^V({3,Graham^X({3,Z,Y}),W}),U}),S}),Q}),O}),M}),K}),I})
	* F=(-(G)kill(☭))^2HIJKLMNOPQRSTUVWXYZ
	* E=10^^F^^G^^H^^I^^J^^K^^L^^M^^N^^O^^P^^Q^^R^^S^^T^^U^^V^^W^^X^^Y^^Z
	* D=10{10{10{10{10{10{10{10{10{10{10{10{10{10{10{10{10{10{10{10{10{E}F}G}H}I}J}K}L}M}N}O}P}Q}R}S}T}U}V}W}X}Y}Z
	* C=Eat(C(1+1/8192),C\*BIGGEST EXTREME ORGANISM,{D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z}) where Eat(a,b,c)=E(a)(a\*b)#^#(X^c[☭] &☭)
	* **B=And(C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z) where And(n,m,@)=And(n&m,@) while preserving the structure and And(n,m)=n&m then solve from left to right using BEAF.**
	* @ = {A{B}C,D{E}F,G{H}I,J{K}L,M{N}O,P{Q}R,S{T}U,V{W}X,Y,Z}
	* ? = ?(@). We have to make a hireachy
		+ ?(1) = x
		+ ?(2) = xx = 49 = Fortnite
		+ ?(3) = x\_x = Death
		+ ?(4) = Deathx = Donut
		+ ?(5) = Deathfortnite = Ligma
		+ ?(6) = my balls
		+ ?(7) = Underwear
		+ ?(8) = Pants
		+ ?(9) = Suit
		+ ?(10) = Camoflauge
		+ ?(11) = Obvious
		+ ?(12) = Idiot
		+ ?(13) = Lottery
		+ ?(14) = Is it a winner?
		+ ?(15) = Bad host
		+ ?(16) = Unknown host
		+ ?(17) = Next season
		+ ?(18) = Unknown season
		+ ?(19) = Next reborn
		+ ?(20) = Unknown reborn
		+ ?(w) = Next VSR
		+ ...
	* > = ?>(?)? where > is chained arrow notation
	* = = > but replace ? with ?(Communism)
	* < = (=^^=)
	* ; = <, but each function is repeated (< but replace every number less than the input with the input) times
	* : = ?(;)
	* 9 = germany: = ANGRY STALIN {;,;,;,;,;,;,;} = {Rayo(:),;,;,;,;,;,;,;,;,;,;}
	* 8 = 9 but replace numbers less than Aleph Null with Aleph Null
	* 7 = 8 but replace numbers less than this "8" with this "8"
	* 6 = 7 but add 999999999999999 to the communist value after each step
	* 5 = 6 but after each step, do {x,6,6,6,6} where 6 was the previous step
	* 4 = 5, but after each step, do Rayo(x), and replace every number smaller than the number of planck times in four's longest screech with the number of planck times in four's longest screech
	* 3 = 4, but after each step, do Illuminati(x)
	* 2 = 3, but all f(x) is replaced with f^x(x)
	* 1 = 2, but all f(a,b,c,d...) is replaced with f(?(a),?(b),?(c),?(d)...)
	* Everything afterward's digits follow these rules
	* / = 11
	* . = 11111111111111111111111111111111111111111111111111111111111111... / times
	* - = 11111111111111111111111111111111111111111111111111111111111111... /\*Communism times
	* , = {-,-,-,-,-,-,-,-,-}
	* + = ,+,
	* \* = ,\*,
	* ) = \*\*\*
	* ( = )))
	* ' = (, but all inputs are (
	* & = previous & previous
	* % = /&100
	* $ = -100% of Communism (all letters, numbers, -, and % follow the rules from earlier)
	* # = {{$,$,$},{$,$,$},{$,$,$}}
	* " = (#\*R\*E\*F)! (once again, the ! is a factorial, and REF follows letter rules)
	* ! = "#REF!" (The ! in the definition is a factorial, and "s are actually #\*R\*E\*F!)
114. C(1+1/2048)=And(**B**((Dab(E(F\_[Previous](Previous))Previous#(F\_[Previous](Previous)),-E(F\_[Previous](Previous))Previous#(F\_[Previous](Previous))))+round(C^-1(Dab(E(F\_[Previous](Previous))Previous#(F\_[Previous](Previous)),-E(F\_[Previous](Previous))Previous#(F\_[Previous](Previous)))))),2,4,8,16,32,64,1024,2048!) where ! is factorial and And is defined on 108.
115. C(1+1/1024)=!="#REF!"
116. C(1.001)=C(1+1/2048)!C(1+1/1024) where ! is Hyperfactorial using HAN
117. C(1.002)=f^%\_O (C(1.001)/(1/1000))
118. C(1.003)=C(1.003)+C(1.002)+C(1.001)+C(1) as a fixed point
119. C(1.005)=C(1.003)\*((2\*%)^^^#)
	* Start with {{$,$,$},{$,$,$},{$,$,$}}
	* Then, do "^^...^^" where A is the previous and there are A arrows
	* (A+N+D)&And(C(1.001),C(1.002),C(1.003))
	* L(C(1.005))^^^AvvvA where v is down arrow notation and L() is defined like this:
		+ L(1)=SANIC(C(1.006)) where C(1.006) is defined like this:
			- SANIC^C(1.005) (1) where 1 is defined above
		+ L(2)=AliA^^^^C(1.007) where C(1.007) is defined as:
			- Start with L(1)
			- Do LEG(L(1)) using the LEG function
			- Add 1332142
			- Multiply by \*
			- Divide by (NEUTRAL-Electrons) and call that \
			- Then, do And() to the following numbers:
				* \
				* A number 6 large=number(6)^2 where number() make the number a string or digit that should be defined in the B section
				* number(9)\*2
				* And a large SODA=SODA^2 where SODA=McDonald(Burger+King) where King is the smallest infinite ordinal
			- After that, square it and you got C(1.007). AliA is defined as Burger\*Mcdonald(King)
		+ L(3)={L(1), **{3,{{$,$,$},{$,$,$},{$,$,$}}(L(2))2}&L** (2)X}& ☭ where the bold is legion array, L(1) is L\*1,X isn't 7, but a structure
		+ L(4)=SCREECH(L(3)+241)^^^SCREECH^AliA (L(3)) where SCREECH(n) is defined as this:
			- Start with n
			- Add +
			- Then, multiply by (\*)
			- Then, do And(Number after previous step,!,",#,$,%)
			- Finally, repeat the previous 4 steps L(3)^n times
		+ L(5)=SANIC(SCREECH(f\_L(4) (L(4))))
		+ L(6)=SCREECH(SANIC(g\_L(4) (L(4))))
		+ L(x) for x greater than or equal to 7 = (Sum from y=1 to y=x-1): E(L(y) & ☭)I#^^#LIE where E is cascading E and I is I and defined above (not L) and LIE is defined as this:
			- L=The L defined above in the alphabet.
			- LI=3\*AliA+({B,O^O(L)M})!
			- LIE=☭{LI}+BiggestSolution($^$) where $ is capitalism
	* Then, do A\*TE where TE=The number of years since 1 BTE when Half Life Never comes out, where Never is defined as this:
		+ N=The N that is defined on the alphabet
		+ NE=Take (N,N) on the coordinate plane. Add 1 to x and 1 to y. Then, sum the x and y.
		+ NEV=How many votes needed to pass through NE lightMATEs of air.
		+ NEVE=The largest naive extension from the function f\_NEV. Then, apply that new function we call H onto NEV to make H(NEV)
		+ NEVER=The limit of R function by Hyp\_cos on to NEVE
	* Then, do A+F where F is defined above
	* (((A)B(A))C((A)B(A)))D(((A)B(A))C((A)B(A)))... Z ... (((A)B(A))C((A)B(A)))D(((A)B(A))C((A)B(A))) where (n)B(m)=f\_n (m) and if we let L[k] be kth letter, then for k>2, (n)L[k](m)=((n)L[k-1](m))L[k-1]((m)L[k-1](n))
	* Finally, (do AliA!(A!(A!(A!(A!))))!BOOM!)![!] using ! defined above, AliA that is also defined, and BOOM is defined as this:
		+ B is B(B) using the B function and the definition of B above
		+ BO=Call O as the Total number of Black Ops 4 players by the release date of Black Ops B. Then, do BodyOdor(B,O) where BodyOdor(n,m)={☭{n},{☭{LI},☭,☭{AliA}},☭{m}}
		+ BOO=BOOM![BO] using a fixed point
		+ BOOM=AliA![BOO![BOOM]] using a fixed point
121. C(1.013)=Unlucky^666 (Lucky^777 (BM\_C(1.008) (999+1))) where Unlucky(n) is defined as:
	* Start with C(1.008) and let A means previous
	* 666A+777n
	* /A\ where /m\ is defined like this:
		+ Start with ☭{☭{☭{m}}}
		+ Add m
		+ Do And(Previous,☭,☭,☭,m)
		+ Take that number, and ☭-ated to the m
		+ Concatenate it to :
		+ Then, Multiply by C(1.008)^C(1.004)
		+ Finally, cube it.
	* /.../n\...\ with A^^^n n's
	* f\_A (n)+%
	* Then, Lucky(n)=/{☭{A},☭{A}(n)☭{A}}\\*And(&,&,&,&,&,...,&) with n^^^A &'s
	* Now, we define Unlucky(n)
	* Start with Lucky(n)
	* Lucky^A (A)
	* 69A+123456789
	* And(A&A,A&A,A&A,☭)
	* NEVER![[BOOM!☭,A],A]
	* Finally, Unlucky(n)=Lucky(Leg(Foot(Rayo(☭{A}))))&n
122. C(1.021)=9+10+Number(9)+Number(10)+C(1.013)
123. C(1.034)=L^Lucky(AliA+Unlucky(C(1.021))) (Unlucky(Lucky(☭{LEG(FOOT(And(C(1.021),C(1.021),C(1.034))))}))) using fixed points
125. C(1.089)=(...((((C(1.055)th prime)th composite)th neither prime nor composite)th square)th cube...)th number that is a number to the C(1.089)^C(1.034) using fixed points
126. C(1.144)=ceil(sqrt((C(1.089)^2)^C(1.1))^3) where C(1.1) is defined as this:
	* !![[C(1.09)^^^C(1.055)]]^#+E(C(1.09))C(1.055)#6123 where C(1.09) is defined as this:
		+ And(((Unlucky(C(1.08)^3)+Lucky(C(1.08)^2))^C(1.08))||C(1.08)![1,1,1,2],-,-,-,-,-) where C(1.08) is defined as this:
			- And(C(1.07)&C(1.089),C(1.07)&C(1.055),C(1.07)&C(1.034),...C(1.07)&C(1.005),C(1.07)&C(1.003),C(1.07)&C(1.002),C(1.07)&C(1.001),C(1.07)&C(1)) where C(1.07) is defined as this:
				* PsiLevel(C(1.06)+Lucky(AliA)^^Elemental C(1.06) (C(1.089))) where C(1.06) is defined as this:
					+ Unlucky^Lucky((C(1.05)\*-+-)^C(1.089)) (A+F![BOOM![C(1.05)],1,C(1.05),C(1.089)]) where C(1.05) is defined as this:
						- [SANIC(SCREECH(]x||C(1.089) (20\*7+-+"^^^C(1.04))...) repeating the functions, where C(1.04) is defined as this:
							* (NEVER!BOOM)!(SANIC(f\_SCREECH(C(1.03)) (C(1.089)))) where C(1.03) is defined as this:
								+ The {C(1.02)<>C(1.089)} fixed point of And(NEVER,BOOM,AliA,C(1.089),C(1.02),C(1.03)) where C(1.02) is defined as:
									- {(C(1.144)^^^^^C(1.02))<>C(1.01)} making it a fixed point, where C(1.01) is defined as:
											+ Item 1 is way bigger that the previous number, Item 0 or C(1.089)
											+ Item 2 is way bigger that the previous number, Item 1
											+ Item 3 is way bigger that the previous number, Item 2
											+ Item 4 is way bigger that the previous number, Item 3
											+ Item 5 is way bigger that the previous number, Item 4
											+ Item 6 is way bigger that the previous number, Item 5
											+ Item 7 is way bigger that the previous number, Item 6
											+ Item 8 is way bigger that the previous number, Item 7
											+ Item 9 is way bigger that the previous number, Item 8
											+ Item 0 is way smaller that the previous number, because we can only afford 1 digit due to budget cuts
											+ Item 1 is way bigger that the previous number, Item 0
											+ Item 2 is way bigger that the previous number, Item 1
											+ Item 3 is way bigger that the previous number, Item 2
											+ Item 4 is way bigger that the previous number, Item 3
											+ Item 5 is way bigger that the previous number, Item 4
											+ Item 6 is way bigger that the previous number, Item 5
											+ Item 7 is way bigger that the previous number, Item 6
											+ Item 8 is way bigger that the previous number, Item 7
											+ Item 9 is way bigger that the previous number, Item 8
											+ Item 0 is way smaller that the previous number, because we can only afford 1 digit due to budget cuts
											+ Item 1 is way bigger that the previous number, Item 0
											+ Item 2 is way bigger that the previous number, Item 1
											+ Item 3 is way bigger that the previous number, Item 2
											+ Item 4 is way bigger that the previous number, Item 3
											+ Item 5 is way bigger that the previous number, Item 4
											+ Item 6 is way bigger that the previous number, Item 5
											+ Item 7 is way bigger that the previous number, Item 6
											+ Item 8 is way bigger that the previous number, Item 7
											+ Item 9 is way bigger that the previous number, Item 8
											+ Item 0 is way smaller that the previous number, because we can only afford 1 digit due to budget cuts
											+ Item 1 is way bigger that the previous number, Item 0
											+ Item 2 is way bigger that the previous number, Item 1
											+ Item 3 is way bigger that the previous number, Item 2
											+ Item 4 is way bigger that the previous number, Item 3
											+ Item 5 is way bigger that the previous number, Item 4
											+ Item 6 is way bigger that the previous number, Item 5
											+ Item 7 is way bigger that the previous number, Item 6
											+ Item 8 is way bigger that the previous number, Item 7
											+ Item 9 is way bigger that the previous number, Item 8
											+ Item 0 is way smaller that the previous number, because we can only afford 1 digit due to budget cuts
											+ Item 1 is way bigger that the previous number, Item 0
											+ Item 2 is way bigger that the previous number, Item 1
											+ Item 3 is way bigger that the previous number, Item 2
											+ Item 4 is way bigger that the previous number, Item 3
											+ Item 5 is way bigger that the previous number, Item 4
											+ Item 6 is way bigger that the previous number, Item 5
											+ Item 7 is way bigger that the previous number, Item 6
											+ Item 8 is way bigger that the previous number, Item 7
											+ Item 9 is way bigger that the previous number, Item 8
											+ Item 0 is way smaller that the previous number, because we can only afford 1 digit due to budget cuts
											+ Item 1 is way bigger that the previous number, Item 0
											+ Item 2 is way bigger that the previous number, Item 1
											+ Item 3 is way bigger that the previous number, Item 2
											+ Item 4 is way bigger that the previous number, Item 3
											+ Item 5 is way bigger that the previous number, Item 4
											+ Item 6 is way bigger that the previous number, Item 5
											+ Item 7 is way bigger that the previous number, Item 6
											+ Item 8 is way bigger that the previous number, Item 7
											+ Item 9 is way bigger that the previous number, Item 8
											+ Item 0 is way smaller that the previous number, because we can only afford 1 digit due to budget cuts
											+ Item 1 is way bigger that the previous number, Item 0
											+ Item 2 is way bigger that the previous number, Item 1
											+ Item 3 is way bigger that the previous number, Item 2
											+ Item 4 is way bigger that the previous number, Item 3
											+ Item 5 is way bigger that the previous number, Item 4
											+ Item 6 is way bigger that the previous number, Item 5
											+ Item 7 is way bigger that the previous number, Item 6
											+ Item 8 is way bigger that the previous number, Item 7
											+ Item 9 is way bigger that the previous number, Item 8
											+ Item 0 is way smaller that the previous number, because we can only afford 1 digit due to budget cuts
											+ Item 1 is way bigger that the previous number, Item 0
											+ Item 2 is way bigger that the previous number, Item 1
											+ Item 3 is way bigger that the previous number, Item 2
											+ Item 4 is way bigger that the previous number, Item 3
											+ Item 5 is way bigger that the previous number, Item 4
											+ Item 6 is way bigger that the previous number, Item 5
											+ Item 7 is way bigger that the previous number, Item 6
											+ Item 8 is way bigger that the previous number, Item 7
											+ Item 9 is way bigger that the previous number, Item 8
											+ Item 0 is way smaller that the previous number, because we can only afford 1 digit due to budget cuts
											+ Item 1 is way bigger that the previous number, Item 0
											+ Item 2 is way bigger that the previous number, Item 1
											+ Item 3 is way bigger that the previous number, Item 2
											+ Item 4 is way bigger that the previous number, Item 3
											+ Item 5 is way bigger that the previous number, Item 4
											+ Item 6 is way bigger that the previous number, Item 5
											+ Item 7 is way bigger that the previous number, Item 6
											+ Item 8 is way bigger that the previous number, Item 7
											+ Item 9 is way bigger that the previous number, Item 8
											+ Item 0 is way smaller that the previous number, because we can only afford 1 digit due to budget cuts
											+ Item 1 is way bigger that the previous number, Item 0
											+ Item 2 is way bigger that the previous number, Item 1
											+ Item 3 is way bigger that the previous number, Item 2
											+ Item 4 is way bigger that the previous number, Item 3
											+ Item 5 is way bigger that the previous number, Item 4
											+ Item 6 is way bigger that the previous number, Item 5
											+ Item 7 is way bigger that the previous number, Item 6
											+ Item 8 is way bigger that the previous number, Item 7
											+ Item 9 is way bigger that the previous number, Item 8
											+ Item 0 is way smaller that the previous number, because we can only afford 1 digit due to budget cuts
127. C(1.233)=Item C(1.144)&C(1.1) using the item function defined above.
128. C(1.250)=C(1.233)+4 (Ends)