Task
Given two sets A and B, compute 



(
A
∖
B
)
∪
(
B
∖
A
)
.


{\displaystyle (A\setminus B)\cup (B\setminus A).}



That is, enumerate the items that are in A or B but not both. This set is called the symmetric difference of A and B.

In other words: 



(
A
∪
B
)
∖
(
A
∩
B
)


{\displaystyle (A\cup B)\setminus (A\cap B)}

 (the set of items that are in at least one of A or B minus the set of items that are in both A and B).

Optionally, give the individual differences (



A
∖
B


{\displaystyle A\setminus B}

 and 



B
∖
A


{\displaystyle B\setminus A}

) as well.



Test cases
A = {John, Bob, Mary, Serena}
B = {Jim, Mary, John, Bob}



Notes
 If your code uses lists of items to represent sets then ensure duplicate items in lists are correctly handled. For example two lists representing sets of a = ["John", "Serena", "Bob", "Mary", "Serena"] and b = ["Jim", "Mary", "John", "Jim", "Bob"] should produce the result of just two strings: ["Serena", "Jim"], in any order.
 In the mathematical notation above A \ B gives the set of items in A that are not in B; A ∪ B gives the set of items in both A and B, (their union); and A ∩ B gives the set of items that are in both A and B (their intersection).


show_sdiff(record u, x)
{
    record r;
    text s;
 
    r.copy(u);
 
    for (s in x) {
        if (r.key(s)) {
            r.delete(s);
        } else {
            r.p_integer(s, 0);
        }
    }
 
    r.vcall(o_, 0, "\n");
}
 
new_set(...)
{
    record r;
 
    ucall(r_p_integer, 1, r, 0);
 
    r;
}
 
main(void)
{
    show_sdiff(new_set("John", "Bob", "Mary", "Serena"),
               new_set("Jim", "Mary", "John", "Bob"));
 
    0;
}