Recall that `nlz(x)` returns the number of leading zeros in the binary representation of `x`. How does this help when `a = 0` or `c = 0` in the `maxor` function?

When `a = 0`, the `maxor` function simplifies to `maxor(0, b, c, d) = (b & ~d) | ((b ^ d) & -(c | d))`. Analyze how `nlz` can optimize the computation of `-(c | d)` in this case.

For the case where `c = 0`, the expression becomes `maxor(a, b, 0, d) = (b & ~d) | ((b ^ d) & -(a | d))`. Use `nlz` to compute `-(a | d)` efficiently by leveraging the fact that `nlz(x) = 32 - floor(log2(x))` for 32-bit integers.