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(void) {
    int n,i,max_f=0,max_s=0,max_f_num,ans=0;
    scanf("%d",&n);
    int c[n],v[n];
    for (i=0; i<n; i++) {
        scanf("%d",&c[i]);
    }
    for (i=0; i<n; i++) {
        scanf("%d",&v[i]);
    }
    //入力終了
    
    for (i=0; i<n; i++) {
        if (max_f < (c[i] - v[i])) {
            max_f = c[i] - v[i];
            max_f_num = i;
        }
    }
    for (i=0; i<n; i++) {
        if (i==max_f_num)
            continue;
        if (max_s < (c[i] - v[i])) {
            max_s = c[i] - v[i];
        }
    }
    ans=max_f+max_s;
    printf("%d\n", ans);
    
}