Wolstenholme's theorem is stated as follows: if $p>3$ is a prime, then \begin{align*} \sum_{k=1}^{p-1}\frac{1}{k}\equiv 0 \pmod{p^2},\\ \sum_{k=1}^{p-1}\frac{1}{k^2} \equiv 0 \pmod{p}. \end{align*} It is also not hard to prove that $$ \sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}\equiv 0 \pmod{p}. $$ However, there are some relationships between $\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}$ and $\sum_{k=1}^{p-1}\frac{1}{k^2}$ mod $p^2$, which I can not prove.

Question: If $p$ is an odd prime, then $$ 4\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}\equiv 3\sum_{k=1}^{p-1}\frac{1}{k^2}\pmod{p^2}. $$ I have verified this congruence for $p$ upto $7919$.

I appreciate any proofs, hints, or references!