Undefined behavior in std::uninitialized_default_construct example on cppreference?

In their example of std::uninitialized_default_construct:

    struct S { std::string m{ "Default value" }; };
 
    constexpr int n {3};
    alignas(alignof(S)) unsigned char mem[n * sizeof(S)];
 
    try
    {
        auto first {reinterpret_cast<S*>(mem)};
        auto last {first + n};  // (1)**********
 
        std::uninitialized_default_construct(first, last);

        // (2)**********
        for (auto it {first}; it != last; ++it) {
            std::cout << it->m << '\n';
        }
 
        std::destroy(first, last);
    }
    catch(...)
    {
        std::cout << "Exception!\n";
    }
 
    // Notice that for "trivial types" the uninitialized_default_construct
    // generally does not zero-fill the given uninitialized memory area.
    int v[] { 1, 2, 3, 4 };
    const int original[] { 1, 2, 3, 4 };
    std::uninitialized_default_construct(std::begin(v), std::end(v));

    // (3)**********
    // for (const int i : v) { std::cout << i << ' '; }
    // Maybe undefined behavior, pending CWG 1997.

I have three questions. (Mark as ********** )

1. Since S and unsigned char aren't pointer-interconvertible, will last point to past-the-end of mem? Is pointer arithmetic undefined behavior here?
2. Is std::launder needed here to make first and last actually point to S? e.g. first = std::launder(first); last = first + n;
3. Is it really undefined behavior to print elements of v? CWG 1997 talk about unsigned char[] storage, but the above example has initialized v with ints.