**Exercise 2.27:** Modify your reverse procedure of exercise 2.18 to produce a `deep-reverse`

procedure that takes a
list as argument and returns as its value the list with its elements reversed and with all sublists deep-reversed
as well. For example,

```
(define x (list (list 1 2) (list 3 4)))
x
((1 2) (3 4))
(reverse x)
((3 4) (1 2))
(deep-reverse x)
((4 3) (2 1))
```

```
(define (deep-reverse x)
(cond ((null? x) '())
((not (pair? x))
(list x))
((not (pair? (car x)))
(append (deep-reverse (cdr x)) (list (car x))))
(else
(append (deep-reverse (cdr x))
(list (deep-reverse (car x)))))))
```

```
> (define x (list (list 1 2) (list 3 4)))
> (deep-reverse x)
((4 3) (2 1))
> (define y (list (list (list 1 2)) (list 3 4)))
> y
(((1 2)) (3 4))
> (deep-reverse y)
((4 3) ((2 1)))
```