No Other Prime Must Divide It

Number Theory Level 4

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 $5$ ), $\omega (100) = 2.$