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: Wrong Answer)

#include <stdio.h>

int main(int argc, char const *argv[]) {
  int a[60], b[60], s[60], count, i, j, tmp;
  scanf("%d", &count);
  for (i = 0; i < count; i++) {
    scanf("%d", &a[i]);
  }
  for (i = 0; i < count; i++) {
    scanf("%d", &b[i]);
  }
  for (i = 0; i < count; i++) {
    s[i] = a[i] - b[i];
  }
  for (i=0; i<count; ++i) {
    for (j=i+1; j<count; ++j) {
      if (s[i] < s[j]) {
        tmp =  s[i];
        s[i] = s[j];
        s[j] = tmp;
      }
    }
  }
  printf("%d\n", s[0] + s[1]);
  return 0;
}