Chance one bubble sort pass fully sorts 100 items

A uniformly random permutation of $1,2,\dots,100$ goes through a single left-to-right pass of bubble sort. The probability that the list is then fully sorted can be written as $\dfrac{a^b}{c!}$ with integers $a,b,c$. Find $a+bc$.

$9902$. Running the pass backwards from the sorted list, each of the $99$ adjacent comparisons is either a swap or not, and each pattern yields a different starting permutation. So $2^{99}$ of the $100!$ permutations work, and $2+99\cdot100=9902$.