(b) Let S(n)logn be a proper complexity function, and let L be a language which is accepted by some S(n)-space bounded Turing machine M. In other words, M never uses more than S(n) space on any input of length n, and M halts and accepts on precisely those inputs x which belong to L; however, for inputs x/L,M may either halt and reject or may run forever. Show that there is a S(n)-space bounded Turing machine M that decides L. In other words, M halts on every input string x and accepts precisely those xL, and never uses more than S(n) space on any input of length n