Inheriting constructor template with access modifiers

Question

When inheriting a constructor template, GCC (6.2.0 via MinGW-w64, 6.3 via Ideone.com) seems to ignore access modifiers.

#include <iostream>

class Base
{
protected:
    template<typename ...Args>
    Base(Args ...args)
    {
        std::cout << "variadic ctor" << std::endl;
    }

    template<typename T>
    Base(T t)
    {
        std::cout << "template ctor" << std::endl;
    }

    Base(char const *s)
    {
        std::cout << s << std::endl;
    }
};

class Derived : public Base
{
protected:
    using Base::Base;
};

int main(int, char**) noexcept
{
    //Base bv{ 0, 1 }; // error: 'Base::Base(Args ...) [with Args = {int, int}]' is protected within this context
    //Base bt{ 0 }; // error: 'Base::Base(T) [with T = int]' is protected within this context
    //Base bs{ "base" }; // error: 'Base::Base(const char*)' is protected within this context
    Derived dv{ 0, 1 };
    Derived dt{ 0 };
    //Derived ds{ "derived" }; // error: 'Derived::Derived(const char*)' is protected within this context
    return 0;
}

None of these should be constructible, but the Derived class is still constructible through the templated constructors. As shown by the char const constructor, only the non-template constructor inherits proper access modifiers.

Live Demo

This is incorrect behavior, isn't it?

EDIT: This answer provides the following related citation when using X::X;:

A constructor so declared has the same access as the corresponding constructor in X.

From 12.9/4, "Inheriting constructors".