[10] A PowerStack consists of a potentially infinite series of stacks So, S1, S2,..., where the i-th stack S, can hold up to 21 elements. In this question, we study the following PowerPush operation for an PowerStack. The main working mechanism of a PowerPush is the following: whenever there is an attempt to Push an element onto any full stack Si, we first Pop all the elements off Si and Push them onto stack Si+1 to make room. If Si+1 is already full, we first recursively move all its members to SH2, etc. Each PowerPush operation starts by attempting to Push to So. Assume that each elementary Push and Pop operation takes 1 unit of time, and the cost of any other operation (e.g., checking if a stack is full) is ignored (a) In the worst case, how many units of time does it take to PowerPush one new element onto a PowerStack containing n elements? Your answer should be in exact form (i.e., not in asymptotic notation) and justified (b) Consider performing a sequence of n PowerPush operations onto a PowerStack that is initially empty. Show that the amortized runtime per operation is O(log n) Hint: You can first try to analyze it using the aggregate method, which is doable but quite complicated; then you should try the accounting method and realize that it becomes much simpler