why pointer to function type is absent in the list of deducible type

template<typename T>
struct Test{};  

template<typename Ret, typename...Args>
struct Test<Ret(*)(Args...)>{
    using type = int;
};
void func(int){}
int main(){
  Test<decltype(&func)>::type b;
}

If type P is Ret(*)(Args...) and the corresponding argument type A is void(*)(int), the deduction attempt to deduce template argument from A, however such form is not listed in the following list which specifies the what form of type can be deduced.

temp.deduct.type#8

A template type argument T, a template template argument TT or a template non-type argument i can be deduced if P and A have one of the following forms:

• T
• cv-list T
• T*
• T&
• T&&
• T[integer-constant]
• template-name (where template-name refers to a class template)
• type(T)
• T()
• T(T)
• T type::*
• type T::*
• T T::*
• T (type::*)()
• type (T::*)()
• type (type::*)(T)
• type (T::*)(T)
• T (type::*)(T)
• T (T::*)()
• T (T::*)(T)
• type[i]
• template-name (where template-name refers to a class template)
• TT
• TT
• TT<>

The list specifies so much however pointer to function type is absent, only function type such as T(T) is comprised, so why a pointer to function type is absent in this list, such type should be deducible.