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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdint.h>
#include <inttypes.h>


int main (int argc, char const *argv[])
{
    int64_t N, V[20], C[20];
    int64_t ans;

    scanf("%"PRId64, &N);
    for (int i = 0; i < N; i++) {
        scanf("%"PRId64, &V[i]);
    }
    for (int i = 0; i < N; i++) {
        scanf("%"PRId64, &C[i]);
    }

    for (int i = 0; i < N; i++) {
        if (V[i] > C[i]) ans += V[i] - C[i];
    }
    
    printf("%"PRId64"\n", ans);
}