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>

int main(){
    int n, i=1;
    char cchar[100], vchar[100], *cptr, *vptr;

    scanf("%d %[^\n] %[^\n]", &n, vchar, cchar);
    // printf("%s \n", vchar);
    // printf("%s \n", cchar);

    int c[n], v[n];
    int ans=0;

    cptr = strtok(cchar, " ");
    c[0] = atoi(cptr);

    while(cptr != NULL) {
        cptr = strtok(NULL, " ");

        if(cptr != NULL) {
            c[i] = atoi(cptr);
        }
        i += 1;
    }

    vptr = strtok(vchar, " ");
    v[0] = atoi(vptr);

    i = 1;
    while(vptr != NULL) {
        vptr = strtok(NULL, " ");

        if(vptr != NULL) {
            v[i] = atoi(vptr);
        }
        i += 1;
    }

    for(i=0; i<n; i++) {
        if(v[i] > c[i]) {
            ans += v[i] - c[i];
            // printf("%d %d %d %d\n", ans, v[i], c[i], i);
        }
    }

    printf("%d\n", ans);

    return 0;
}