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 <bits/stdc++.h>
using namespace std;
#define rep(i, n)     for(int i = 0; i < (int)(n); i++)
#define rep2(i, m, n) for(int i = m; i < (int)(n); i++)
#define MOD 1000000007

int main() {
  int N;
  cin >> N;
  
  vector<int> Vi,Ci; 
  rep(i,N){
    int tmpVi;
    cin >> tmpVi;
    Vi.push_back(tmpVi); 
  }
  
  rep(i,N){
    int tmpCi;
    cin >> tmpCi;
    Ci.push_back(tmpCi); 
  }
  
  int ans = 0;
  rep(i,N){
    int earn = Vi.at(i) - Ci.at(i);
    if(earn > 0) ans += earn;
  }
  cout << ans << endl;
  return 0;
}