Primes

Let $p = 11$ , which seems to satisfy the condition, else $p$ is disordered with $p + 11$ . In the second case, $p + 11\vert p(p+1)(p+2)$ if it also divides the product $(p+1)(p+2)$ . Which $= (-10)(-9)$ in $\mod{p + 11}$ , so it must be true that $p + 11\vert90$ , so $p\in {7, 19, 79}$

Therefore $7 + 11 + 19 + 79 = \boxed{116}$