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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<algorithm>
#include <functional>

int main(void){
    int N, C[50], V[50];
    scanf("%d", &N);
    for(int i=0;i<N;i++){
        scanf("%d", &V[i]);
    };
    for(int i=0;i<N;i++){
        scanf("%d", &C[i]);
    }
    int vc[50];
    for(int i=0;i<N;i++){
        vc[i] = V[i] - C[i];
    }
    std::sort(vc, vc+N, std::greater<int>());
    int sum = 0;
    for(int i=0;i<N;i++){
        // printf("%d ", vc[i]);
        if(vc[i] > 0) sum += vc[i];
        else break;
    }
    printf("%d", sum);
}