Template argument deduction for member function pointers

It is known that template arguments can be pointers to member functions.

So I can write:

struct Bar
{
    int fun(float x);
};

template <int (Bar::*FUN)(float)>
struct Foo
{ /*...*/ };

typedef Foo<&Bar::fun> FooBar;

But what if I want the the Bar type itself to be a template argument:

template <typename B, int (B::*FUN)(float)>
struct Foo
{ /*...*/ };

typedef Foo<Bar, &Bar::fun> FooBar;

Now, when I use it, I have to write Bar twice!

My question is: Is there a way to force the compiler to deduce the class type automatically?

The objective is for this to just work:

typedef Foo<&Bar::fun> FooBar;
typedef Foo<&Moo::fun> FooMoo;