Specializations that are structurally identical

Consider the two partial specializations below:

#include <iostream>
#include <vector>
#include <type_traits>

template <typename, typename...> struct A;

template <typename... Ts>
struct A<int, Ts...> {
    void foo (int a) const {std::cout << a << '\n';}
    void operator()(const std::vector<int>& v) const {std::cout << v.size() << '\n';}
};

template <typename... Ts>
struct A<char, Ts...> {
    void foo (char a) const {std::cout << a << '\n';}
    void operator()(const std::vector<char>& v) const {std::cout << v.size() << '\n';}
};

int main() {
    A<int, long, double> a;
    A<char, float, bool, short> b;
    a.foo(5);  // 5
    b.foo('!');  // !
    a({1,2,3});  // 3
    b({1,2,3});  // 3
}

How to write the two specializations just once?

template <typename T, typename... Ts>
struct A<T, Ts...> {
    static_assert (std::is_same<T,int>::value || std::is_same<T,char>::value, "Error");
    void foo (T a) const {std::cout << a << '\n';}
    void operator()(const std::vector<T>& v) const {std::cout << v.size() << '\n';}
};

doesn't work because it doesn't specialize anything, and I can't place class = std::enable_if<std::is_same<T,int>::value || std::is_same<T,char>::value, T>::type anywhere because a default argument cannot go after a pack. The above specializations shall only be for int and char. Any other type will some other general definition for the class.