Egyptian equation I

The equation $\frac{1}{x}+\frac{1}{y}=\frac{1}{n}$ with $0<x\le y$ has $14$ solutions for $n=12345$:

• $\frac{1}{12346}+\frac{1}{152411370}=\frac{1}{12345}$
• $\frac{1}{12348}+\frac{1}{50812020}=\frac{1}{12345}$
• $\frac{1}{12350}+\frac{1}{30492150}=\frac{1}{12345}$
• $\frac{1}{12354}+\frac{1}{16945570}=\frac{1}{12345}$
• $\frac{1}{12360}+\frac{1}{10172280}=\frac{1}{12345}$
• $\frac{1}{12370}+\frac{1}{6108306}=\frac{1}{12345}$
• $\frac{1}{12390}+\frac{1}{3398990}=\frac{1}{12345}$
• $\frac{1}{12420}+\frac{1}{2044332}=\frac{1}{12345}$
• $\frac{1}{12570}+\frac{1}{689674}=\frac{1}{12345}$
• $\frac{1}{13168}+\frac{1}{197520}=\frac{1}{12345}$
• $\frac{1}{14814}+\frac{1}{74070}=\frac{1}{12345}$
• $\frac{1}{16460}+\frac{1}{49380}=\frac{1}{12345}$
• $\frac{1}{19752}+\frac{1}{32920}=\frac{1}{12345}$
• $\frac{1}{24690}+\frac{1}{24690}=\frac{1}{12345}$

How many solutions are there for $n=2091129587$ ?

Answer format: $count,\sum x,\sum y$

Example: 14,200392,273453272 for n = 12345

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