Detect if expression result is constexpr

Suppose I have some template class A which takes an integer parameter:

template<int N>
struct A;

Now I would like to build a factory method which builds an A based on some evaluation of a function f, which may or may not be constexpr. If the evaluated result is constexpr, the return value should be A<f(args...)>. Otherwise, I want the return result to be A<0>.

Conceptually, I want something like this:

template<class F, class... Args>
auto factory(F&& f, Args&&... args)
  -> (is_result_constexpr(f(std::forward<Args>(args)...)) ? A<f(std::forward<Args>(args)...)> : A<0>)
{
    // ... dispatch to appropriate helper to construct constexpr/non-constexpr versions
}

• How would I do this (is this even possible)?
• If this isn't possible in C++11, what about C++14/C++17?

As an example usage of what I want, suppose I have a constexpr function f0:

constexpr int f0(int N)
{
    return N;
}

int main()
{
    auto a0 = factory(f0, 1); // should have type A<1>
    int val;
    std::cin >> val; // val isn't compile time known
    auto a1 = factory(f0, val); // should have type A<0>
}