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 <cstdio>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <vector>
#include <map>
#include <stack>
#include <queue>
#include <string>
#include <algorithm>
#include <functional>
#include <utility>
#include <numeric>
#include <set>

using namespace std;
using llong = long long;
const int MOD=1000000007;

int main(int argc, char**argv){
    int ans = 0;
    int n;
    cin >> n;
    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];
    }

    vector<int> diff(n);
    for (int i = 0; i < n; i++){
        diff[i] = v[i] - c[i];
    }
    for (int i = 0; i < n; i++){
        if (diff[i] > 0) ans += diff[i];
    }

    cout << ans << endl;




    return 0;
}