Conjugacy and the Class Equation · Superchallenge problem

$A_5$ (order $60$) is simple: it has no nontrivial proper normal subgroup. If some non-identity element had a centralizer of index $2$ (order $30$), that centralizer would automatically be normal (any subgroup of index $2$ is), contradicting simplicity. Given $A_5$'s actual class equation $1+12+12+15+20=60$, what IS the largest centralizer order among its non-identity elements?