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;

#ifndef DEBUG
#define dump(...)
#endif

signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    size_t N;
    cin >> N;

    vector<int> value(N), cost(N);
    for (auto &vi : value) cin >> vi;
    for (auto &ci : cost) cin >> ci;

    int res = 0;
    for (size_t i = 0; i < N; ++i) {
        int diff = value[i] - cost[i];
        if (diff > 0) res += diff;
    }

    cout << res << "\n";

    return 0;
}