Consider the consecutive primes $p_1 = 19$ and $p_2 = 23$. It can be verified that $1219$ is the smallest number such that the last digits are formed by $p_1$ whilst also being divisible by $p_2$.

In fact, with the exception of $p_1 = 3$ and $p_2 = 5$, for every pair of consecutive primes, $p_2 \gt p_1$, there exist values of $n$ for which the last digits are formed by $p_1$ and $n$ is divisible by $p_2$. Let $S$ be the smallest of these values of $n$.

Find $\sum S$ for every pair of consecutive primes with $5 \le p_1 \le 1000000$.