Super Pow - Modular Exponentiation Rules

Remember the modulo property: $(a \times b) \mod m = ((a \mod m) \times (b \mod m)) \mod m$. This applies to $(a^{12})^{10} \times a^3$ as well.

You compute $a^{12} \mod 1337$ first (recursively). Then raise that result to the 10th power, modulo 1337. Multiply by $a^3 \mod 1337$, and take modulo again.

Each exponentiation step uses the fast power function. The recursion depth is at most the number of digits in $b$, which keeps the computation manageable.