A, B, C, D are four points on a circle. AC and BD intersect at a point such that $$\angle{BEC} = 130^o$$ and $$\angle{ECD} = 20^o$$. Then, $$\angle{BAC}$$ is

A

90 $$^{\circ}$$

B

100 $$^{\circ}$$

C

110 $$^{\circ}$$

D

120 $$^{\circ}$$