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 <vector>

using namespace std;

using ull = unsigned long long;

int main() {
  ull n;
  cin >> n;

  std::vector<ull> v(n);
  std::vector<ull> c(n);

  for (ull i = 0; i < n; i++) cin >> v[i];
  for (ull i = 0; i < n; i++) cin >> c[i];
  
  ull m = 1 << n;
  int mval = 0;
  
  for (ull i = 0; i < m; i++) {
    int l = 0;
    ull i_ = i, k = 0;
    while (i_) {
      if (i_ & 0x01) {
	l = l + v[k] - c[k];
      }
      
      i_ = i_ >> 1;
      k++;
    }

    if (i) {
      mval = max(mval, l);
    } else {
      mval = l;
    }
  }

  cout << mval;
  
  return 0;
}