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110. Diophantine reciprocals II

In the following equation x, y, and n are positive integers.

$${\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{y}}={\displaystyle \frac{1}{n}}$$

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.