Please help with this OCaml code asap!!

Note that function calls have the highest precedence in OCaml, so the last expression is the same as (f00f(n/2))+(f00f(n/2)). If we assume that the function f can compute its result in constant time and constant space, what are the (worst-case) time and space complexities of the function foo? Justify your answer. (It doesn't have to be a full mathematical proof, but it should be a convincing argument.) Assume that the integer input n is always non-negative, and assume the usual applicative-order evaluation rule