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 ll = long long;
using Pii = std::pair<int, int>;
using Pll = std::pair<ll, ll>;

const ll INF = (ll)1e11;

using namespace std;

int main() {
    ios::sync_with_stdio(false);

    int N;
    cin >> N;

    vector<int> C, V;
    for (int i = 0, v; i < N; ++i) {
        cin >> v;
        V.push_back(v);
    }
    for (int i = 0, c; i < N; ++i) {
        cin >> c;
        C.push_back(c);
    }

    int ans = -1e8;
    for (unsigned s = 0; s < (1 << N); ++s) {
        int X = 0, Y = 0;
        for (int i = 0; i < N; ++i) {
            if ((s >> i) & 1) {
                X += V[i];
                Y += C[i];
            }
        }
        ans = max(ans, X - Y);
    }
    
    cout << ans << endl;

    return 0;
}