Problem:
Score : 200 points

Problem Statement

There are N gems. The value of the i-th gem is V_i.

You will choose some of these gems, possibly all or none, and get them.

However, you need to pay a cost of C_i to get the i-th gem.

Let X be the sum of the values of the gems obtained, and Y be the sum of the costs paid.

Find the maximum possible value of X-Y.

Constraints

All values in input are integers.

1 \leq N \leq 20

1 \leq C_i, V_i \leq 50

Input

Input is given from Standard Input in the following format:

N
V_1 V_2 ... V_N
C_1 C_2 ... C_N

Output

Print the maximum possible value of X-Y.

Sample Input 1

3
10 2 5
6 3 4

Sample Output 1

5

If we choose the first and third gems, X = 10 + 5 = 15 and Y = 6 + 4 = 10.
We have X-Y = 5 here, which is the maximum possible value.

Sample Input 2

4
13 21 6 19
11 30 6 15

Sample Output 2

6

Sample Input 3

1
1
50

Sample Output 3

0

Solution in Common Lisp (SBCL 1.1.14) (judge verdict: Accepted)

#-swank
(unless (member :child-sbcl *features*)
  (quit
   :unix-status
   (process-exit-code
    (run-program *runtime-pathname*
                 `("--control-stack-size" "128MB"
                   "--noinform" "--disable-ldb" "--lose-on-corruption" "--end-runtime-options"
                   "--eval" "(push :child-sbcl *features*)"
                   "--script" ,(namestring *load-pathname*))
                 :output t :error t :input t))))

(defun split (x str)
  (let ((pos (search x str))
        (size (length x)))
    (if pos
      (cons (subseq str 0 pos)
            (split x (subseq str (+ pos size))))
      (list str))))

(setq *n* (read))
(setq *v* (make-array *n* :initial-contents (mapcar #'parse-integer (split " " (read-line)))))
(setq *c* (make-array *n* :initial-contents (mapcar #'parse-integer (split " " (read-line)))))

(setq *gains* nil)

(defun dfs (gain i)
  (let ((next-gain gain) (result nil))
      (if (>= i *n*)
        (setq *gains* (cons gain *gains*))
        (dotimes (n 2 result)
                 ;(print (list "next-gain " next-gain (aref *v* i) (aref *c* i)))
                 (when (= n 1) (setq next-gain (+ gain (- (aref *v* i) (aref *c* i)))))
                 (setq result (dfs next-gain (1+ i)))))))

  (dfs 0 0)
  (print (apply #'max *gains*))