$3+8=4+7=5+6=11$ 
There is a cube whose six faces are marked with $3$, $4$, $5$, $6$, $7$, and $8$. If the sum of every two numbers that are on the opposite faces are the same, the number on the opposite face of $4$ is~\uline{~~~~~~~~~~}~ 
Options:
(A) $$3$$ 
(B) $$5$$ 
(C) $$6$$ 
(D) $$7$$ 
(E) $$8$$ 