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 N=0;
   puts("Nを入力");
   scanf("%d",&N);
   int V[N];
   int C[N];
   int X,Y;
   int i=0;
   for(int k=0;k<N;k++){
       printf("V%dを入力してください\n",k+1);
       scanf("%d",&V[k]);
       printf("C%dを入力してください\n",k+1);
       scanf("%d",&C[k]);
   }
       X=V[0]+V[N-1];
       Y=C[0]+C[N-1];
        printf("%d\n",(X-Y));
        return 0;    
   }