$3+8=4+7=5+6=11$ 
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 identical, the number on the opposite face of $4$ is~\uline{~~~~~~~~~~}~ 

a: $$3$$ 
b: $$5$$ 
c: $$6$$ 
d: $$7$$ 
e: $$8$$ 