# Why does Carter-Wegman MAC have 2n-bit output?

The Coursera course on Cryptography 1 has a section about Carter-Wegman MAC. Here is the image of slide:

Now my question is both $F$ and $S$ output $n$-bit strings. We XOR two $n$-bit strings and we should get an $n$-bit string as a result. But the slide says the output result is a $2n$-bit string. How is this possible?

The MAC consists of the pair of the n-bit random value $r$ and the n-bit value $t = F(k_1,r) \oplus S(k_2,m)$. The length of $(r,t)$ is at least $2n$ bits, partly depending on how it is encoded. If the two values are simply concatenated, the total length is $2n$, as stipulated.
• I forgot the point that the sender should send that random number r. – Majid Azimi May 11 '16 at 9:39