Five random intervals sharing a common point

Each of five intervals is produced by drawing two independent $U[0,1]$ points and using the smaller as the left end and the larger as the right end. The probability that all five intervals have a point in common is $\tfrac pq$ in lowest terms. Find $p+q$.

$71$. Only the relative order of the ten endpoints matters. The intervals share a point exactly when every left end precedes every right end; that happens with probability $\tfrac{(5!)^2\,2^5}{10!}=\tfrac{8}{63}$, so $p+q=71$.