AHSME 1999 Problem 25

There are unique integers $a_{2},a_{3},a_{4},a_{5},a_{6},a_{7}$ such that $$\frac {5}{7} = \frac {a_{2}}{2!} + \frac {a_{3}}{3!} + \frac {a_{4}}{4!} + \frac {a_{5}}{5!} + \frac {a_{6}}{6!} + \frac {a_{7}}{7!}$$ where $0\leq a_{i} < i$ for $i = 2,3,\ldots,7$. Find $a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}$. $\textrm{(A)} \ 8 \textrm{(B)} \ 9 \textrm{(C)} \ 10 \textrm{(D)} \ 11 \textrm{(E)} \ 12$