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 <string>
#include <vector>
#include <list>
#include <algorithm>
#include <numeric>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <functional>
#include <utility>
#include <cctype>
#include <fstream>
#include <cstring>

using namespace std;

#define ll long long
#define MOD 1000000007
#define INF 2000000000
#define LINF 9000000000000000000
#define rep(i,n) for(int i=0;i<(int)n;i++)
#define du(x) cout<<#x<<'='<<x<<endl
#define du2(x,y) cout<<#x<<'='<<x<<' '<<#y<<'='<<y<<'\n'<<endl
#define du3(x,y,z) cout<<#x<<'='<<x<<' '<<#y<<'='<<y<<' '<<#z<<'='<<z<<'\n'<<endl
#define darr(arr,n) rep(i,(n-arr)){cout<<*(arr+i)<<' ';}cout<<endl
#define ddarr(arr,a,b) rep(i,a){rep(j,b){cout<<*(arr+i*b+j)<<' ';}cout<<endl;}
#define ddd ifstream in("Text.txt");cin.rdbuf(in.rdbuf());

int V[50], C[50];
int main() {
    

    int N;
    cin >> N;
    rep(i, N) {
        cin >> V[i];
    }
    rep(i, N)cin >> C[i];

    int ma = 0;
    int X = 0, Y = 0;
    rep(i, N) {
        if (V[i] > C[i]) {
            X += V[i];
            Y += C[i];
        }

    }

    

    cout << X-Y << endl;

}