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: Wrong Answer)

#include <bits/stdc++.h>
#define INF (1<<30)
#define MOD 1000000007
#define l_ength size
using ll = long long;
using namespace std;

int main(){
	int n; cin >> n;
	vector<int> v(20, 0);
	vector<int> c(20, 0);
	for( int i = 0; i < n; ++i ) cin >> v[i];
	for( int i = 0; i < n; ++i ) cin >> c[i];
		int ans = 0;
	for( int i = 0; i < ((1<<20)-1); ++i ){
		int x = 0, y = 0;
		for( int j = 0; j < 20; ++j ){
			if( i & (1<<j) ){
				x += v[j];
				y += c[j];
			}
		}
		ans = max( ans, x-y );
	}
	cout << ans << endl;
	return 0;
}