For $p$ a prime larger than $2k+2$, is it possible that $p^2$ divides $B_{2k}$?

Question

One characterization of the irregular primes is as follows:

For an irregular prime $p$, there exists some natural number $k\le \frac{p-3}{2}$ such that $p$ divides [the numerator of] $B_{2k}$ where $B_{2k}$ is a Bernoulli number.

Question: Is there some prime $p$ for which there exists some natural number $k\le \frac{p-3}{2}$, such that $p^2$ divides $B_{2k}$?

References that provide an answer, or just deal with the question are welcome.