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 <algorithm>
#include <cassert>
#include <cctype>
#include <climits>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstring>
#include <deque>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <string>
#include <tuple>
#include <vector>
#define rep(i, n) for (int i = 0; i < (int)(n); ++i)
//#define cerr if(false) cerr
#ifdef DEBUG
#define show(...) cerr << #__VA_ARGS__ << " = ", debug(__VA_ARGS__);
#else
#define show(...) 42
#endif
using namespace std;
using ll = long long;
using pii = pair<int, int>;
template <typename T, typename S>
ostream &operator<<(ostream &os, pair<T, S> a) {
    os << '(' << a.first << ',' << a.second << ')';
    return os;
}
template <typename T>
ostream &operator<<(ostream &os, vector<T> v) {
    for (auto x : v) os << x << ' ';
    return os;
}
void debug() {
    cerr << '\n';
}
template <typename H, typename... T>
void debug(H a, T... b) {
    cerr << a;
    if (sizeof...(b)) cerr << ", ";
    debug(b...);
}
int main(){
    int n;
    cin >> n;
    vector<int>a(n),b(n);
    rep(i,n)cin >> a[i];
    rep(i,n)cin >> b[i];
    int ans = 0;
    for(int i = 0; i < (1<<n); i++){
        int x = 0, y = 0;
        for(int j = 0; j < n; j++){
            if(i & (1<< j)){
                x += a[j];
                y += b[j];
            }
        }
        ans = max(ans, x - y);
    }
    cout << ans << endl;
}