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 <iostream>
#include <algorithm>
#include <climits>
#include <numeric>
#include <iomanip>
#include <cmath>
#include <vector>
#include <set>
#include <unordered_set>
#include <stack>
#include <map>

#define pii pair<int, int>

using namespace std;

bool comp(pii a, pii b){
    return a.first - a.second < b.first - b.second;
}

int main() {
    int n;
    cin >> n;
    vector <pii> a(n);
    for(int i = 0; i < n; i++) cin >> a[i].first;
    for(int j = 0; j < n; j++) cin >> a[j].second;
    sort(begin(a), end(a), comp);
    
    int ans = 0;
    
    for(int i = 0; i < n; i++){
        //cout << a[i].first << ' ' << a[i].second << endl;
        if(a[i].first > a[i].second) ans += a[i].first - a[i].second;
    }
    cout << ans;
}