Eccoci.

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

$$\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{n}$$

For $n = 4$ there are exactly three distinct solutions:

$$\begin{align} \dfrac{1}{5} + \dfrac{1}{20} &= \dfrac{1}{4}\\ \dfrac{1}{6} + \dfrac{1}{12} &= \dfrac{1}{4}\\ \dfrac{1}{8} + \dfrac{1}{8} &= \dfrac{1}{4} \end{align} $$

What is the least value of $n$ for which the number of distinct solutions exceeds one-thousand?

NOTE: This problem is an easier version of Problem 110; it is strongly advised that you solve this one first.