If :$\sum_{k=1}^{n}k^p =(\frac{n(n+1)}{2})\mod(p)$ how i deduce the remain of :$\sum_{k=1}^{n}k^{-p}$?

Question

I have tried to determine the remain of this serie:$\sum_{k=1}^{n}k^p$ :

I got this formula $\sum_{k=1}^{n}k^p =(\frac{n(n+1)}{2})\mod(p)$ ,where $p$

is prime and $k$ is positive integer .Now my question is :if what i got

is true then how do I deduce the remain of this :$\sum_{k=1}^{n}k^{-p}$

using $\sum_{k=1}^{n}k^p =(\frac{n(n+1)}{2})\mod(p)$?

note: $k^{-p}$ is inverse of $k$ modulo $p $ and I work in $\mathbb{Z_{p}}$

Any kind of help is appreciated.Thank you