Having the output size be the same as the input size (i.e., \( o = i \)) can be a desirable property:
For any \( i \) and for \( k \) odd (\( k = 2n + 1, \quad n \in \mathbb{N} \)), \( s = 1 \) and \( p = \lfloor k / 2 \rfloor = n \),
$$ \begin{equation*} \begin{split} o &= i + 2 \lfloor k / 2 \rfloor - (k - 1) \\ &= i + 2n - 2n \\ &= i. \end{split} \end{equation*} $$