Is it possible to get a (member) function pointer to a specific instantiation of a generic lambda?

I know I can do so for standard non capturing lambdas, and for abbreviated templates, but I can't seem to be able to get a member function pointer for the explicitly instantiated operator() call operator member function of the invented type for the generic lambda.

#include <iostream>
 
void f1( auto v)  { std::cout << v << std::endl; }
int main() {
    void (*pf)(int) = f1<int>; // OK
    void (*pf2)(int) = [](int v)  { std::cout << v << std::endl; } ; // OK
    [](auto v) { std::cout << v << std::endl; }.operator() < int > (42); // OK
    auto generic_template = [](auto v)  { std::cout << v << std::endl; } ;
    using generic_type = decltype (generic_template);
    // void (generic_type::*pf3)(int) = &generic_type::operator()<int>;  // fails to compile

    pf(5);
}

The interest here is academic.

Edit:

As a note of interest to future readers the solutions offered to this question also apply to getting function pointers for lambdas with capture, in addition to generic lambdas. For example, based on the answers :

  auto generic_lambda = [](auto v)  { std::cout << v << std::endl; } ;
  using generic_type = decltype (generic_lambda);
  void (generic_type::*pf1)(int) const = &generic_type::operator();
  (&generic_lambda->*pf1)(43); // OK

  int x = 5;
  auto capturing_lambda = [x](int v)  { std::cout << v+x << std::endl; } ;
  using capturing_type = decltype (capturing_lambda);
  void (capturing_type::*pf2)(int) const = &capturing_type::operator();
  (&capturing_lambda->*pf2)(43); // OK