Miner's Arrow Number

Miner's Arrow Number is so big that it is (7000↑↑↑↑↑↑↑↑↑↑78)*0% of the size of the Omniverse and bigger than the "OH NO" number!
Sadly, it doesn't end in 120 steps.
This is the formula:
Ma_0 = 2
Ma_1 = Ma_0 \uparrow \uparrow Ma_0
Ma_2 = Ma_1 \overbrace{\uparrow ... \uparrow}^{ Ma_1 \uparrow} Ma_1
Ma_3 = Ma_2 \overbrace{\uparrow ... \uparrow}^{ Ma_2 \uparrow} Ma_2
⋮
Ma_{118} = Ma_{117} \overbrace{\uparrow ... \uparrow}^{ Ma_{117} \uparrow}Ma_{117}
Ma_{119} = Ma_{118} \overbrace{\uparrow ... \uparrow}^{ Ma_{118} \uparrow} Ma_{118}
Ma_{120} = Ma_{119} \overbrace{\uparrow ... \uparrow}^{ Ma_{119} \uparrow} Ma_{119}
Ma_{121} = Ma_{120} \overbrace{\uparrow ... \uparrow}^{ Ma_{120} \uparrow} Ma_{120}
Ma_n = Ma_{n-1} \overbrace{\uparrow ... \uparrow}^{ Ma_{n-1} \uparrow}Ma_{n-1}
Conclusion: It is extremely big, and I know and have proof it is bigger than Graham's number. And it's value is Ma_{121}.
== Proof Ma121 is bigger than Graham's Number ==
In the second step of Mas, it is 4↑↑↑↑4, which is bigger than g1 (first step of Graham's Number), but as the number of steps is bigger than G, and the up arrow numbers change (G's numbers never change, as it is three). That means it is bigger than G. Is it bigger than TREE(3)? I actually don't know. (No its not because TREE(3) is a legiattic array)
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