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

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

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

$$\begin{array}{rl}{\displaystyle \frac{1}{5}}+{\displaystyle \frac{1}{20}}& ={\displaystyle \frac{1}{4}}\\ {\displaystyle \frac{1}{6}}+{\displaystyle \frac{1}{12}}& ={\displaystyle \frac{1}{4}}\\ {\displaystyle \frac{1}{8}}+{\displaystyle \frac{1}{8}}& ={\displaystyle \frac{1}{4}}\end{array}$$

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.