--description--

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, $p_2 > p_1$, 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 $5 ≤ p_1 ≤ 1000000$.

--hints--

primePairConnection() should return 18613426663617120.

assert.strictEqual(primePairConnection(), 18613426663617120);

--seed--

--seed-contents--

function primePairConnection() {

  return true;
}

primePairConnection();

--solutions--

// solution required