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: Wrong Answer)

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();

        int[] v = new int[n];
        int[] c = new int[n];

        for (int i = 0; i < n; i++) {
            v[i] = sc.nextInt();
        }

        for (int i = 0; i < n; i++) {
            c[i] = sc.nextInt();
        }

        int max = 0;

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (max < ((v[i] + v[j]) - (c[i] + c[j])) && i != j) {
                    max = ((v[i] + v[j]) - (c[i] + c[j]));
                }
            }
        }

        System.out.println(max);
    }
}