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)

//
//  main.cpp
//  ABC125
//
//  Created by hiroaki on 2019/04/27.
//  Copyright © 2019年 hiroaki. All rights reserved.
//

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>

using namespace std;

int main(){
//    int a,b,t;
//    cin >>a>>b>>t;
//    int s=(t+0.5)/a;
//    cout<<s*b<<endl;
    
    int n = 0;
    cin >> n;
    vector<int> v(n),c(n);
    
    for (int i=0; i<n; i++) {
        cin >> v.at(i);
    }
    for (int i=0; i<n; i++) {
        cin >> c.at(i);
    }
    vector<int> dif; // 差
    
    for (int i=0; i<n; i++) {
        dif.push_back(v[i]-c[i]);
    }
    
    sort(dif.begin(), dif.end());
    
    int difference=0;
    for (int i=0; i<n; i++) {
        if(dif[n-1-i]<0){
            break;
        }
        difference+=dif[n-1-i];
    }
    cout<<difference<<endl;
    return 0;
}