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: Wrong Answer)

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int main(){
    int N; std::cin >> N;
    std::vector<int>V;
    std::vector<int>C;
    int ref;
    for (int i = 0; i < N; i++)
    {
        std::cin >> ref;
        V.push_back(ref);
    }
    
    for (int i = 0; i < N; i++)
    {
        std::cin >> ref;
        C.push_back(ref);
    }

    int Ans = 0;
    std::vector<int>diff;
    for (int i = 0; i < N; i++)
    {
        ref = V[i] - C[i];
        //cout<<ref<<endl;
        diff.push_back(ref);
    }
    std::sort(diff.begin(),diff.end());
    //cout << diff[0] << ":"<<diff[1] << ":"<<diff[2] << ":"<<endl;
    int tmp = 0;
    for (int i = N; i>0  ; --i)
    {
       // cout<<diff[i]<<endl;
        tmp += diff[i];
        if(Ans <= tmp)Ans = tmp;

    }
    std::cout << Ans << std::endl;
    return 0;
}