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.io.PrintWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.Scanner;

public class Main {
    public static void main(String args[]) {
        Scanner scan = new Scanner(System.in);
        List<Integer> list = new ArrayList<>();
        List<Integer> list2 = new ArrayList<>();
        int count = 0;
        int a = Integer.parseInt(scan.next());

        for(int i = 0;i<a;i++){
            list.add(Integer.parseInt(scan.next()));
        }
        for(int i = 0;i<a;i++){
            list2.add(Integer.parseInt(scan.next()));
        }
        for(int i = 0;i<a;i++){
            if(list.get(i) - list2.get(i) > 0){
              count += (list.get(i) - list2.get(i));
            }
        }
       PrintWriter out = new PrintWriter(System.out);
        out.println(count);
        out.flush();
        scan.close();
    }
}