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: Runtime Error)

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

int main() {
    int n;
    int v[n], c[n];
    scanf("%d", n);

    char str0[n], *ch0;
    fgets(str0, n, stdin);
    char str1[n], *ch1;

    ch0 = strtok(str0, " \n");
    ch1 = strtok(str1, " \n");
    for (int i = 0; i < n; ++i) {
        if (ch0 == NULL)
            break;
        else
            v[i] = atoi(ch0);
        ch0 = strtok(NULL, " \n");
        if (ch1 == NULL)
            break;
        else
            c[i] = atoi(ch1);
        ch1 = strtok(NULL, " \n");
    }

    int dif;
    int sum = 0;
    for (int k = 0; k < n; ++k) {
        if (v[k] > c[k]){
            dif = v[k] - c[k];
            sum += dif;
        }
    }

    printf("%d", sum);

    return 0;
}