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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
int main(){
int n,mx=0;
vector<pair<int,int> >v;
cin>>n;
int a[n+1],b[n+1];
for(int i=0;i<n;i++){
    cin>>a[i];
}
for(int i=0;i<n;i++){
    cin>>b[i];
    mx=max(mx,a[i]-b[i]);
}
for(int i=0;i<n;i++){
    v.push_back(make_pair(a[i],b[i]));
}
sort(v.begin(),v.end());
int sum=0;
for(int i=0;i<n;i++){
    sum+=(v[i].first-v[i].second);
    mx=max(mx,sum);
}
reverse(v.begin(),v.end());
sum=0;
for(int i=0;i<n;i++){
    sum+=(v[i].first-v[i].second);
    mx=max(mx,sum);
}
cout<<mx;
}