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 <string.h>
#include <stdlib.h>
#include <math.h>

int main() {

    int n, v, c;
    scanf("%d", &n);

    int arr01[n];
    int arr02[n];

    int mysum = 0;

    for (int i = 0; i < n; i++) {
        scanf("%d", &v);
        arr01[i] = v;

    }
    for (int j = 0; j < n; j++) {
        scanf("%d", &c);
        arr02[j] = c;
    }
    for (int k = 0; k < n; k++) {

        if (arr01[k] - arr02[k] > 0) {
            mysum += arr01[k] - arr02[k];
        }

    }

    printf("%d", mysum);


    return 0;

}