The solutions to the equations $z^2 = 4 + 4\sqrt{15}i$ and $z^2 = 2 + 2\sqrt{3}i$, where $i = \sqrt{-1}$, form the vertices of a parallelogram in the complex plane. The area of this parallelogram can be written in the form $p\sqrt{q} - r\sqrt{s}$, where $p, q, r,$ and $s$ are positive integers and neither $q$ nor $s$ is divisible by the square of any prime number. What is $p + q + r + s$?
方程 $z^2 = 4 + 4\sqrt{15}i$ 和 $z^2 = 2 + 2\sqrt{3}i$ 的解(其中 $i = \sqrt{-1}$)在复平面中形成一个平行四边形的顶点。这个平行四边形的面积可以写成 $p\sqrt{q} - r\sqrt{s}$ 的形式,其中 $p, q, r, s$ 是正整数,且 $q$ 和 $s$ 都不被任何质数的平方整除。求 $p + q + r + s$ 的值。