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)

/*input
1
1
50
*/
#include <bits/stdc++.h>
using namespace std;

#define pb push_back
#define f first
#define s second
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define int long long
int const mod = 1e9+7;
int power(int x, int y, int mod = 2e18){ int ans = 1;x %= mod;while(y){if(y&1)ans = (x * ans) % mod;x = (x * x) % mod;y >>= 1;}return ans;}
int gcd(int a , int b){if(a==0)return b;return gcd(b%a,a);}
int a[21];
int b[21];
int n;
int final_ans = 0 ;
void solve(int  i,int j,int k){
  if(i == n){
    final_ans = max(final_ans,j-k);
    return;
  }
  solve(i+1,j+a[i],k+b[i]);
  solve(i+1,j,k);
}
signed main() {
  IOS;
  
  cin>>n;
  
  for(int i=0;i<n;i++)cin>>a[i];
  for(int i=0;i<n;i++)cin>>b[i];
 solve(0,0,0);
  cout << final_ans;
return 0;
}