What is the running time of the following code: * 1 123 & 4 5678 9 def bs1(L, item): 0(1) O O(logn) O O(n) OO(nlogn) O O(n^2) O 0(2^n) O(n!) if len (L) == 0: return False median= len (L) // 2 if item == L[median]: return True elif item < L[median]: 1 point return bs1(L[:median], item) return bs1(L[median+ 1:], item) else: