A parallel plate capacitor of capacitance $$12.5 \mathrm{~pF}$$ is charged by a battery connected between its plates to potential difference of $$12.0 \mathrm{~V}$$. The battery is now disconnected and a dielectric slab $$(\epsilon_{\mathrm{r}}=6)$$ is inserted between the plates. The change in its potential energy after inserting the dielectric slab is ________ $$\times10^{-12} \mathrm{~J}$$.