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>

#define rep(x, y) for (int i = (x); i < (y); i++)

typedef long long ll;

using namespace std;

void solve() {
    int n;
    cin >> n;

    int v[n], c[n];
    rep(0, n) cin >> v[i];
    rep(0, n) cin >> c[i];

    int x, y, diff = 0;
    for (int bit = 0; bit < (1<<n); bit++) {
        vector<int> select;
        for (int i = 0; i < n; i++) {
            if (bit & (1<<i)) {
                select.push_back(i);
            }
        }

        x = y = 0;
        for (int s : select) {
            x += v[s];
            y += c[s];
        }

        diff = max(diff, x - y);
    }

    cout << diff << endl;
}

int main() {
    solve();

    return 0;
}