Consider unordered lists wherein elements may occur multiple times. (10 Points) For example, A = [5, 3, 7, 5, 5, 1, 2, 10, 3, 3, 3, 2, 3, 4, 7, 1, 1] Write a Python function frequencyCount(A) that takes an unordered list A as input and returns a list of pairs (a, n), where a is an element in A that occurs n times. Your algorithm should have a time complexity of O(|A| log2 |A|). So, applied to the above list, frequencyCount(A) should return the list: [(1, 3),(2, 2),(3, 5),(4, 1),(5, 3),(7, 2),(10, 1)]