For all \(n \in \mathbb{Z},\) let \(\omega (n) \) denote the number of distinct prime divisors of \(n.\) Find the sum of all primes \(p\) such that \(\omega ((p-1)!+1) = 1.\)

**Details and assumptions**

- As an explicit example, since \(100= 2^2 \times 5^2\) has two distinct prime divisors (\(2\) and \(5\)), \(\omega (100) = 2.\)

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