smart pointer to const in function's signature

I was wondering if the implicit cast I have when passing a shared_ptr < T>& as argument to a function taking shared_ptr < const T>& involves some hidden costs (such as the construction of a copy).

void f(shared_ptr<const T>& )

main{
   auto p = make_shared<Widget>(); 
   f(p);
   return 0;
}

Similarly I was wondering if I have an extra cost when passing shared_ptr < T> instead of shared_ptr < const T>. (I'm assuming that in both cases I am paying for the refcount increment and decrement).