# Add two rational numbers... esoterically

## Objective
Build a binary operator \$\star : \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}\$ such that, there exists a bijection \$i : \mathbb{Z} \to \mathbb{Q}\$ such that, for every integer \$M\$ and \$N\$, \$M \star N = i^{-1}(i(M) + i(N))\$ holds.

Or in mathematical terms, build a binary operator \$\star\$ that admits a group isomorphism from \$(\mathbb{Z}, \star)\$ to \$(\mathbb{Q}, +)\$.

## I/O Format
Flexible. Standard loopholes apply.

## Scoring
This is a [tag:fastest-algorithm] challenge. The submission whose the **asymptotic** worst-case time complexity of \$\star\$ is the fastest wins.

The parameter for the time complexity is \$s(M) + s(N)\$, where \$s\$ is defined as follows:

* \$s(0) = s(-1) = 0\$.
* For every integer \$D > 0\$, \$s(D) = \lfloor \log_2 D \rfloor + 1\$.
* For every integer \$D < -1\$, \$s(D) = \lfloor \log_2 (-D-1) \rfloor + 1\$.

The shortness of your code is a tiebreaker. That is, this challenge is secondarily [tag:code-golf].