How to get a forwarding function to call a base template function that is declared after the forwarding function

I have a case where I need to have a forwarding function defined before a template base function is defined/declared. However, if I call the forwarding function (fwd) that in turn calls the base function test, it says that the base template function is not visible (see the code below). However, if test is called directly, everything works.

So my question is this, is it possible to have the forwarding function make a call to a base template function that is defined later in the compilation unit (before it is used but after the forwarding function)? If not, do I have any options to work around this? I would like to avoid a forward declaration before fwd as that would make use of the library I am developing harder. I think if I could force fwd to be inline it would solve the problem but I have no way of doing that unless a macro is used.

#include <iostream>
#include <vector>

template<typename T, std::enable_if_t<std::is_scalar<T>::value, int> = 0>
void test(const T& t)
{
  std::cout << "Scalar"  << std::endl;
}

template<typename T>
void fwd(T&& t)
{
  test(std::forward<T>(t));
}

template<typename T>
void test(const std::vector<std::vector<T>>& t)
{
  std::cout << "vector vector of T" << std::endl;
}

int main(int argc, const char * argv[]) {

  test(1);  //OK, prints Scalar
  fwd(1);   //OK, prints Scalar

  test(std::vector<std::vector<int>>()); //OK, prints vector vector of T

  // Causes compile error: Call to function 'test' that is neither visible in the template definition
  // nor found by argument dependent lookup
  fwd(std::vector<std::vector<int>>());

  return 0;
}