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>
#include<stdlib.h>
#include<string.h>
int main(void){
  int n;
  int i,j;

  scanf("%d",&n);
  int v[n];
  int c[n];
  int rv[n];
  int tmp;
  int ans=0;
  for(i=0;i<n;i++){
    scanf("%d",&v[i]);
  }
  for(i=0;i<n;i++){
    scanf("%d",&c[i]);
  }
  for(i=0;i<n;i++){
    rv[i]=v[i]-c[i];
  }
  for(i=0;i<n;i++){
    for(j=i+1;j<n;j++){
      if(rv[i]<rv[j]){
        tmp=rv[j];
        rv[j]=rv[i];
        rv[i]=tmp;
      }
    }
  }
  for(i=0;i<n;i++){
    if(rv[i]>=0){
      ans=ans+rv[i];
    }
  }
  printf("%d",ans);

return 0;
}