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552. Chinese leftovers II

Let A_n be the smallest positive integer satisfying A_n mod p_i = i for all 1 ≤ i ≤ n, where p_i is the i-th prime.
For example A₂ = 5, since this is the smallest positive solution of the system of equations

• A₂ mod 2 = 1
• A₂ mod 3 = 2

The system of equations for A₃ adds another constraint. That is, A₃ is the smallest positive solution of

• A₃ mod 2 = 1
• A₃ mod 3 = 2
• A₃ mod 5 = 3

and hence A₃ = 23. Similarly, one gets A₄ = 53 and A₅ = 1523.

Let S(n) be the sum of all primes up to n that divide at least one element in the sequence A.
For example, S(50) = 69 = 5 + 23 + 41, since 5 divides A₂, 23 divides A₃ and 41 divides A₁₀ = 5765999453. No other prime number up to 50 divides an element in A.

Find S(300000).