Number of divisors of the form $(4n+1)$ 
Find the number of divisors of $$2^2\cdot3^3\cdot5^3\cdot7^5$$ which are of the form $(4n+1)$  

I know how to find the total number of divisors. But, to find the number of divisors of the form $(4n+1)$, I'm thinking of listing down the divisors and then finding, but that'd be very tedious. Is there any elegant way to do this?  
Any help will be appreciated.
Thanks.

Number of divisors of $$N= 2^2\times3^3\times5^3\times7^5,$$which are of the form $4n+1$ exculuding 
$$\begin{align}1 &=\{ \text{number of terms in product}\}\\
&=(1+3^2)(1+5+5^2+5^3)(1+7^2+7^4)+\{\text{number of terms in product}\}\\
&= ( 3+3^3)(7+7^3+7^5)(1+5+5^2+5^3)-1\\
&= 2\times4\times3+2\times3\times4-1\\
&= 47
\end{align}$$
