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 (Clang 3.8.0) (judge verdict: Accepted)

#include<stdio.h>
#include<string.h>
int max(int a,int b){
  if(a > b)return a;
  return b;
}
int min(int a,int b){
  if(a < b)return a;
  return b;
}
int zet(int a){
  if(a > 0)return a;
  return -a;
}
void swap(int *x,int *y) {
  int temp = *x;
  *x = *y;
  *y = temp;
}
int partition(int array[],int left,int right) {
  int i,j,pivot;
  i = left;
  j = right + 1;
  pivot = left;
  do{
    do{ i++; } while(array[i] < array[pivot]);
    do{ j--; } while(array[pivot] < array[j]);
    if(i < j)swap(&array[i],&array[j]);
  }while (i < j);
  swap(&array[pivot],&array[j]);
  return j;
}
void quick_sort(int array[],int left,int right) {
  int pivot;
  if(left < right) {
    pivot = partition(array,left,right);
    quick_sort(array,left,pivot-1);
    quick_sort(array,pivot+1,right);
  }
}
int main(void){
  int n,cnt = 0;
  scanf("%d",&n);
  int a[n],b,c;
  for(int i = 0;i < n;i++){
    scanf("%d",&a[i]);
  }
  for(int i = 0;i < n;i++){
    scanf("%d",&b);
    c = a[i] - b;
    if(c > 0){
      cnt += c;
    }
  }
  printf("%d\n",cnt);
  return 0;
}