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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;

class Main {
  public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);
    int num = Integer.parseInt(scanner.nextLine());
    String[] value = scanner.nextLine().split(" ");
    String[] cost = scanner.nextLine().split(" ");
    
    ArrayList list = new ArrayList<Integer>();
    int tmp = 0;
    for (int i = 0; i < num; i++) {
      if ((tmp = Integer.parseInt(value[i]) - Integer.parseInt(cost[i])) > 0) {
        list.add(tmp);
      }
    }
    if (list.size() == 0) {
      System.out.println("0");
    }
    else {
      Collections.sort(list, Comparator.reverseOrder());
      int sum = 0;
      for (int i = 0; i < list.size(); i++) {
        sum += (Integer)list.get(i);
      }
      System.out.println(sum);
    }
  }
}