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.*;
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();
    }
    
    System.out.println(maxValue(v, c, 0));
    
    //System.out.println(Arrays.toString(v));
    //System.out.println(Arrays.toString(c));
  }
  
  private static int maxValue(int[] v, int[] c, int start) {
    
    if (start == v.length - 1) {
      return Math.max(v[start] - c[start], 0);
    }
    
    int value1 = v[start] - c[start];
    int value2 = maxValue(v, c, start + 1);
    int sum = value1 + value2;
    
    return Math.max(sum, value2);
  }
}