Function Template Overloading with Different Return Types

Following code snippets are from https://en.cppreference.com/w/cpp/language/function_template#Function_template_overloading.

How is it possible to overload functions/function templates with return type A<I+J> Vs A<I-J>?

Did the page really mean that overload #1 & overload #2 compose valid function overload set?

Or I misunderstood the meaning of the page?

template<int I, int J>
A<I+J> f(A<I>, A<J>); // overload #1
 
template<int K, int L>
A<K+L> f(A<K>, A<L>); // same as #1
 
template<int I, int J>
A<I-J> f(A<I>, A<J>); // overload #2

I have tried followings in godbolt.org which did not compile as expected with errors:

ASM generation compiler returned: 1
<source>: In function 'int main()':
<source>:48:24: error: call of overloaded 'func<1, 2>(A<1>, A<2>)' is ambiguous
   48 |     A<-1> a = func<1,2>(A<1>{}, A<2>{});
      |               ~~~~~~~~~^~~~~~~~~~~~~~~~
<source>:35:8: note: candidate: 'A<(I + J)> func(A<I>, A<J>) [with int I = 1; int J = 2]'
   35 | A<I+J> func(A<I>, A<J>) {
      |        ^~~~
<source>:41:8: note: candidate: 'A<(I - J)> func(A<I>, A<J>) [with int I = 1; int J = 2]'
   41 | A<I-J> func(A<I>, A<J>) {
      |        ^~~~

template <int>
struct A {
};

template <int I, int J>
A<I+J> func(A<I>, A<J>) {
    std::cout << "func1\n";
    return A<I+J>{};
}

template <int I, int J>
A<I-J> func(A<I>, A<J>) {
    std::cout << "func2\n";
    return A<I-J>{};
}

int main()
{
    A<-1> a = func<1,2>(A<1>{}, A<2>{});
    return 0;
}