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 "bits/stdc++.h"

using namespace std;

using ll = long long;
using ld = long double;
using P = pair<int, int>;
constexpr ld EPS = 1e-12;
constexpr int INF = numeric_limits<int>::max() / 2;
constexpr int MOD = 1e9 + 7;

template <typename T>
void printv(const vector<T> &v)
{
    int sz = v.size();
    for (int i = 0; i < sz; i++)
    {
        cout << v[i] << " \n"[i == sz - 1];
    }
}

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);

    int n;
    cin >> n;
    int ret = 0;
    vector<int> v(n), c(n);
    for (int i = 0; i < n; i++)
    {
        cin >> v[i];
    }
    for (int i = 0; i < n; i++)
    {
        cin >> c[i];
        ret += max(0, v[i] - c[i]);
    }
    cout << ret << endl;
}