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 Java7 (OpenJDK 1.7.0) (judge verdict: Accepted)

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.Scanner;

public class Main {
	public static void main(String[]args) {
		Scanner scan = new Scanner(System.in);
		int n = scan.nextInt();
		int count = 0 ;
	    int arr[] = new int[n];
	    int arr2[] = new int[n];
	    for(int  i = 0 ; i < n ; i ++) {
	    	arr[i] = scan.nextInt();
	    }
	    for(int  i = 0 ; i < n ; i ++) {
	    	arr2[i] = scan.nextInt();
	    }
	    int sum = 0 ;
	    for(int i = 0 ; i < n ; i++) {
	    	if(arr[i] - arr2[i] >0) {
	    		sum+= arr[i] - arr2[i];
	    	}
	    }
	    System.out.println(sum);
	}
}