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>

#define HMax 100 //配列の上限？
#define YES puts("Yes");
#define NO puts("No");

int fs(int count, int*);//1種の数列入力(数列) scanf("%d", &n); n = fs(a); n = 要素数 a = 数列
int fs_2(int*, int*);//2種の数列入力(数列) n = fs(a, b); n = 要素数 a = 数列

int main(void) {
  int a[HMax], b[HMax], n, i, s[HMax], ans, max = 0, max2 = 0, flag;
  scanf("%d", &n);
  fs(n, a);
  fs(n, b);
  for (i = 0; i < n; i++) {
    s[i] = a[i] - b[i];
    if(s[i] > max) {
      max = s[i];
      flag = i;
    }
  }
  for (i = 0; i < n; i++) {
    if(s[i] > max2 && i != flag) {
      max2 = s[i];
    }
  }
  ans = max + max2;
  printf("%d\n", ans);
  return 0;
}

int fs(int count,int *suretu) {
  int i;
  for (i = 0; i < count; i++) {
    scanf("%d", &suretu[i]);
  }
  return count;
}

int fs_2(int *suretu, int *suretu_2) {
  int i, count;
  scanf("%d", &count);
  for (i = 0; i < count; i++) {
    scanf("%d %d", &suretu[i], &suretu_2[i]);
  }
  return count;
}