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 C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>

void maximum_profit(int *C, int *V, int N, int index, int profit, int *max_profit) {
  if (index < N) {
    maximum_profit(C,V,N,index+1,profit+V[index]-C[index],max_profit);
    maximum_profit(C,V,N,index+1,profit,max_profit);
  } else if (profit > (*max_profit)) {
    (*max_profit) = profit;
  }
}

void Resale(int N) {
  int C[N], V[N], i, max_profit;
  for (i = 0; i < N && scanf("%d",&V[i]); i++);
  for (i = 0, max_profit = 0; i < N && scanf("%d",&C[i]); i++);
  maximum_profit(C,V,N,0,0,&max_profit);
  printf("%d",max_profit);
}

int main(void) {
  int N;
  scanf("%d",&N);
  Resale(N);
  return 0;
}