Miner's Arrow Number

This 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)