2009 Fermat Problem 16

Six dice are stacked on the floor as shown. On each die, the 1 is opposite the 6, the 2 is opposite the 5, and the 3 is opposite the 4. What is the maximum possible sum of numbers on the 21 visible faces?

\textbf{(A)}\ 69\quad \textbf{(B)}\ 88\quad \textbf{(C)}\ 89\quad \textbf{(D)}\ 91\quad \textbf{(E)}\ 96

If there are no answer choices shown, enter a numerical answer.

Full credit to this problem is given to the CEMC, you may view all Fermat contests here.