Specialize member const resolution of template

Having guard class on C++11 that is responsible to invoke some member function on scope exit:

template <class T, void (T::*op)()>
struct Guard
{
    Guard(T*g):
        _g(g){}
    ~Guard()
    {
        (_g->*op)();
    }
    T*_g;
};

Usage is very simple:

typedef Guard<Foo, &Foo::bar> FooGuard;
...
FooGuard g(&foo);

My question originated from existing shared_ptr<Foo>. How to create specialization that keeps shared_ptr<T> instead of T*

What I've already tried:

template <class T, void (T::*op)()>
struct Guard<std::shared_ptr<T>, op>
{
    Guard(std::shared_ptr<T>& g):
        _g(g){}
    ~Guard()
    {
        ((*_g).*op)();
    }

    std::shared_ptr<T> _g;
};

But during compilation on G<std::shared_ptr<Foo>, &Foo::bar> g2(foo); have foreseeable got:

error C2440: 'specialization' : cannot convert from 'overloaded-function' to 'void (__thiscall std::shared_ptr::* )(void)'