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>
#include <stdlib.h>
#include <math.h>
#include <limits.h>
#include <string.h>
#define ll long long int

int main()
{
    int N;
    scanf("%d",&N);
    int V[N];
    int C[N];
    for (int i=0; i<N; i++)
        scanf("%d",&V[i]);


    for (int i=0; i<N; i++)
        scanf("%d",&C[i]);
    int max = 0;
    int sum = 0;
    int X,Y;
    for (int j=0; j<N; j++)
    {
        for (int i=j+1; i<N; i++)
        {
            X=V[j]+V[i];
            Y=C[j]+C[i];
           sum=X-Y;
           if (max<=sum)max=sum;
        }
    }
    printf("%d",max);



    return 0;
}