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[] Values = new int[N];
		int[] Costs = new int[N];
		int max=0;

		for(int i=0;i<N;i++) {
			Values[i]=sc.nextInt();
		}
		for(int i=0;i<N;i++) {
			Costs[i]=sc.nextInt();
		}

		for(int i=0;i<N;i++) {
			int dif=Values[i]-Costs[i];
			if(max<dif) {
				max=dif;
			}
			for(int j=i+1;j<N;j++) {
				int dif2=Values[j]-Costs[j];
				dif2+=dif;
				if(max<dif2) {
					max=dif2;
				}
			}
		}
		System.out.println(max);
	}
}