为什么从引用生成共享指针会复制对象?

我打算让构造函数接受引用,然后创建指向作为引用传递的对象的指针,并将这些指针存储在字段中。然而,由于某种原因,我这样做的方式正在创建副本,我不明白为什么:

#include <iostream>
#include <vector>
#include <memory>

// the class with the field of pointers
template<class T, class... Types>
class C
{

private:
    std::vector<std::shared_ptr<T>> mem; // the field of pointers

public:
    C(T& t, Types&... args)
      // make pointers to the objects passed by reference and store them in mem
      : mem{ std::make_shared<T>(t), std::make_shared<T>(args)... }
    {

        // to demonstrate that the pointers point to copies, alter one of the objects (subscript operator expected)
        (*mem[0])[0] = 10;


        // demonstrate the effect on the local copies
        std::cout << "vectors in mem:" << "\n";
        for (const auto& x : mem) {
            for (const auto& y : *x) {
                std::cout << y << ' ';
            }
            std::cout << "\n";
        }
    }
};

int main()
{
    std::vector<int> v1{ 1, 2, 3 };
    std::vector<int> v2{ 1, 2, 3 };
    std::vector<int> v3{ 1, 2, 3 };

    // make an object of type C with some vectors to store pointers to in field mem
    C<std::vector<int>, std::vector<int>, std::vector<int>> c(v1, v2, v3);

    // demonstrate that original vectors are unaltered
    std::cout << "original vectors:"<< "\n";

    for (const auto& y : v1) {
        std::cout << y << ' ';
    }
    std::cout << "\n";

    for (const auto& y : v2) {
        std::cout << y << ' ';
    }
    std::cout << "\n";

    for (const auto& y : v3) {
        std::cout << y << ' ';
    }
    std::cout << "\n";
}

我错过了什么?复制发生在哪里?为什么?