Diophantine Reciprocals II
In the following equation $x$, $y$, and $n$ are positive integers.
$$\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{n}$$
It can be verified that when $n = 1260$ there are $113$ distinct solutions and this is the least value of $n$ for which the total number of distinct solutions exceeds one hundred.
What is the least value of $n$ for which the number of distinct solutions exceeds four million?
NOTE: This problem is a much more difficult version of Problem 108 and as it is well beyond the limitations of a brute force approach it requires a clever implementation.