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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