Powers of Two

$2^7=128$ is the first power of two whose leading digits are "$12$". The next power of two whose leading digits are "$12$" is $80$.

Define $p(L, n)$ to be the $n$th-smallest value of $j$ such that the base $10$ representation of $2^j$ begins with the digits of $L$. So $p(12, 1)=7$ and $p(12, 2)=80$.

You are also given that $p(123, 45) = 12710$.

Find $p(123, 678910)$.