the sum \[\sum_{n=1}^\infty \dfrac{\tau(n)}{n^2}=\dfrac{\pi^4}{36}\] This has a very simple closed form, and is easy to work out.

However, the sum: \[\sum_{n=1}^\infty \dfrac{\ln(n)\tau(n)}{n^2}\] Has a very difficult closed form, of the form \[-\dfrac{\pi^a}{b}(c\gamma+\ln(d\pi)+f\ln(A))\] Find \(a+b+c+d+f\).

**Notations:**

All of \(a,b,c,d,f\) are integers.

\(\tau(n)\) means the number of positive divisors of n.

A is the Glaisher-Kinkelin Constant

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