Can we use "using ..." with the name of the parameter package?

Question

Suppose we have the following “overloaded lambdas template” for two versions (taken from here ):

template <class F1, class F2>
struct overload_set : F1, F2
{
    overload_set(F1 x1, F2 x2) : F1(x1), F2(x2) {}
    using F1::operator();
    using F2::operator();
};

template <class F1, class F2>
overload_set<F1,F2> overload(F1 x1, F2 x2)
{
    return overload_set<F1,F2>(x1,x2);
} 

which we can use like this:

auto f = overload (
    []( int) { cout << __PRETTY_FUNCTION__ << endl; }, 
    [](char) { cout << __PRETTY_FUNCTION__ << endl; }
);

f('k');
f( 2 );

Now the question is, can we take a general approach to this? I tried with the following (implementation from the same site):

template <class... F>
struct overload_set : F... 
{
  overload_set(F... f) : F(f)... {}  
};

template <class... F>
auto overload(F... f) 
{
  return overload_set<F...>(f...);   
}

but then I get:

main.cpp: In function ‘int main()’:
main.cpp:59:6: error: request for member ‘operator()’ is ambiguous
   f(1);
      ^
main.cpp:51:14: note: candidates are: main()::<lambda(char)>
       [](char) { cout << __FILE__ << ":" << __LINE__ << ":" << __PRETTY_FUNCTION__ << endl; });
              ^
main.cpp:50:13: note:                 main()::<lambda(int)>
       [](int) { cout << __FILE__ << ":" << __LINE__ << ":" << __PRETTY_FUNCTION__ << endl; },
             ^
main.cpp:60:8: error: request for member ‘operator()’ is ambiguous
   f('a');
        ^
main.cpp:51:14: note: candidates are: main()::<lambda(char)>
       [](char) { cout << __FILE__ << ":" << __LINE__ << ":" << __PRETTY_FUNCTION__ << endl; });
              ^
main.cpp:50:13: note:                 main()::<lambda(int)>
       [](int) { cout << __FILE__ << ":" << __LINE__ << ":" << __PRETTY_FUNCTION__ << endl; },