What is the purpose of this trailing return type declaration?

Question

I am trying to understand the example provided at cppreference for std::visit. Here is the example from that link:

    #include <iomanip>
    #include <iostream>
    #include <string>
    #include <type_traits>
    #include <variant>
    #include <vector>


    template<class T> struct always_false : std::false_type {};

    using var_t = std::variant<int, long, double, std::string>;

    template<class... Ts> struct overloaded : Ts... { using Ts::operator()...; };
    template<class... Ts> overloaded(Ts...) -> overloaded<Ts...>;

    int main() {
        std::vector<var_t> vec = {10, 15l, 1.5, "hello"};
        for(auto& v: vec) {
            // void visitor, only called for side-effects
            std::visit([](auto&& arg){std::cout << arg;}, v);

            // value-returning visitor. A common idiom is to return another variant
            var_t w = std::visit([](auto&& arg) -> var_t {return arg + arg;}, v);

            std::cout << ". After doubling, variant holds ";
            // type-matching visitor: can also be a class with 4 overloaded operator()'s
            std::visit([](auto&& arg) {
                using T = std::decay_t<decltype(arg)>;
                if constexpr (std::is_same_v<T, int>)
                    std::cout << "int with value " << arg << '\n';
                else if constexpr (std::is_same_v<T, long>)
                    std::cout << "long with value " << arg << '\n';
                else if constexpr (std::is_same_v<T, double>)
                    std::cout << "double with value " << arg << '\n';
                else if constexpr (std::is_same_v<T, std::string>)
                    std::cout << "std::string with value " << std::quoted(arg) << '\n';
                else 
                    static_assert(always_false<T>::value, "non-exhaustive visitor!");
            }, w);
        }

        for (auto& v: vec) {
            std::visit(overloaded {
                [](auto arg) { std::cout << arg << ' '; },
                [](double arg) { std::cout << std::fixed << arg << ' '; },
                [](const std::string& arg) { std::cout << std::quoted(arg) << ' '; },
            }, v);
        }
    }

I am particularly interested in the overloaded declarations:

    template<class... Ts> struct overloaded : Ts... { using Ts::operator()...; };
    template<class... Ts> overloaded(Ts...) -> overloaded<Ts...>;

The first of these makes sense - the struct overloaded is being declared with constructors for each type it encapsulates. However I don't understand the purpose of the second. This seems to be a function declaration similar to:

template<class... Ts> overloaded<Ts...> overloaded(Ts...);

But the function is never defined, so how can it be used later in the example? If this is just a constructor then why doesn't it need an overloaded:: prefix and where is the body? I think I am just misunderstanding what the trailing return type declaration "expands" into so any insight into what is accomplished by the overloaded(Ts...) declaration would be appreciated.