Consider a hash function $h$ that maps numerous keys into an array $A$, indexed from 0 to $n-1$. The concern is that excessive collisions arising from the hashed keys can potentially cause inefficiency within the hash table. The following pseudocode, while not complete, implements a function that utilizes the divide-and-conquer strategy to identify the case where strictly more than half of the keys share the same hash value. The function takes the hash function $h$ and the subarray $A$ with the index range from $low$ to $high$ as input and returns the majority hash value shared by over $\lfloor (high-low+1)/2 \rfloor$ keys. If the keys stored in the subarray $A[low..high]$ do not have such a majority hash value, the function returns $-1$.

1: function FindMajorityHashValue($h$, $A$, $low$, $high$) 2: if $low = high$ then return $h(A[low])$ end if 3: $mid \leftarrow \lfloor (low+high)/2 \rfloor$ 4: $leftMajority \leftarrow$ FindMajorityHashValue($h$, $A$, $low$, $mid$) 5: $rightMajority \leftarrow$ FindMajorityHashValue($h$, $A$, $mid+1$, $high$) 6: $leftCount \leftarrow$ CountElement($A$, $low$, $high$, $leftMajority$) 7: $rightCount \leftarrow$ CountElement($A$, $low$, $high$, $rightMajority$) 8: [待填空] 9: end function

In the pseudocode, CountElement($A$, $low$, $high$, $val$) calculates and returns the count of keys in the subarray $A[low..high]$ whose hashed values are equal to the given $val$. Please select the correct description(s) below.