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 Ma_s, 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)