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.*;
import java.util.stream.*;
import java.util.function.*;

class Main {
  public static void main(String[] args){
    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    Supplier<IntStream> repn=  ()-> IntStream.range(0,n);
//    IntStream repn = IntStream.range(0,n);
    Integer[] a =repn.get().map(x-> sc.nextInt())
                           .boxed()
                           .toArray((i) -> new Integer[i]);
    Integer[] b =repn.get().map(x-> sc.nextInt())
                           .boxed()
                           .toArray((i) -> new Integer[i]);
    int value =  repn.get().map(i-> a[i]-b[i])
                           .filter(x-> x>0)
                           .sum();
    System.out.println(value);
}}