Consider the consecutive primes p₁ = 19 and p₂ = 23. It can be verified that 1219 is the smallest number such that the last digits are formed by p₁ whilst also being divisible by p₂.

In fact, with the exception of p₁ = 3 and p₂ = 5, for every pair of consecutive primes, p₂ p₁, there exist values of n for which the last digits are formed by p₁ and n is divisible by p₂. Let S be the smallest of these values of n.

Find S for every pair of consecutive primes with 5 p₁ 1000000.