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>

using namespace std;

#define MAX_N 20
#define MAX_V 50
#define MAX_C 50

int V[MAX_V], C[MAX_C];

int main( int argc, char **argv )
{
    int N;
    cin >> N;

    for ( int i = 0; i < N; i++ ) cin >> V[i];
    for ( int i = 0; i < N; i++ ) cin >> C[i];

    int max_diff = 0;
    for ( int i = 0; i < ( 1 << N ); i++ )
    {
        int diff = 0;
        for ( int j = 0; j < N; j++ )
            if ( i & ( 1 << j ) ) diff += V[j] - C[j];
        max_diff = max( max_diff, diff );
    }

    cout << max_diff << endl;

    return 0;
}