Maya lists all the positive divisors of $2010^2$. She then randomly selects two distinct divisors from this list. Let $p$ be the [probability](https://artofproblemsolving.com/wiki/index.php/Probability) that exactly one of the selected divisors is a [perfect square](https://artofproblemsolving.com/wiki/index.php/Perfect_square). The probability $p$ can be expressed in the form $\frac {m}{n}$, where $m$ and $n$ are [relatively prime](https://artofproblemsolving.com/wiki/index.php/Relatively_prime) positive integers. Find $m + n$.
$2010^2 = 2^2\cdot3^2\cdot5^2\cdot67^2$. Thus there are $(2+1)^4$ divisors, $(1+1)^4$ of which are squares (the exponent of each prime factor must either be $0$ or $2$). Therefore the probability is
\[\frac {2\cdot2^4\cdot(3^4 - 2^4)}{3^4(3^4 - 1)} = \frac {26}{81} \Longrightarrow 26+ 81 = \boxed{107}.\]