Consider the prune-and-search algorithm using the median of medians for finding the $k$-th smallest element in an unsorted array $A$ of $n$ distinct elements, where $1 \le k \le n$, as described below:

Algorithm Select(A, k): Input: An array A of n elements and an integer k, $1 \le k \le |A| = n$ Output: The k-th smallest element in A Step 1: If $|A| \le 5$, then sort array A (with any method, e.g., insertion sort) and return the element with rank k (i.e., the k-th element in the sorted array). Step 2: Divide the array A into groups of five elements each (ignore the last group if it has fewer than five elements). Step 3: Find the median of each group. Step 4: Recursively compute the median of the found medians, denoted by m. Step 5: Partition the original array into the following three sets: $L = \{x \in A \mid x < m\}$ $E = \{m\}$ $R = \{x \in A \mid x > m\}$ Step 6: If $k = |L| + 1$ then return $m$ Else if $k \le |L|$ then return Select(L, k) Else return Select(R, k - |L| - 1).

Select one or more correct statements.