In $$\triangle ABC $$, AB and AC are produced to points D and E, respectively. If the bisectors of $$ \angle CBD$$ and $$ \angle BCE $$ meet at the point O, and $$\angle BOC= 57^\circ$$, then $$ \angle A$$ is equals to:

A

$$93^\circ$$

B

$$66^\circ$$

C

$$114^\circ$$

D

$$57^\circ$$