Finite group with at least 3 generators? The answer to this question could be trivial, but usually I do not work in group theory. 

Is there a  finite nonabelian group $G$ that is generated by $S$ where $S$ is a minimal generating set of $G$, and $3\leq |S|\leq |G|-1$? (Other than the direct product of nonabelian or abelian groups)

What I'm looking here is finite nonabelian group that satisfies the above inquality. Of course
the obvious answer is the direct product of certain finite nonabelian or abelian  group, but I'm  not interested in it.
For the infinite nonabelian groups we can take the free group of n-generators (with n≥3).
I notice most of the finite nonabelian groups  like dihedral, symmetric group,.. does not satisfy the above condition.
Any help will be useful!

What about $\;Q_8\times C_2\;,\;\;Q_8=$ the quaternion group?
