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"
using namespace std;
#define LL long long
#define REP(i, n) for(int (i)=0; (i)<(n); (i)++)
#define PB push_back
#define MP make_pair
#define all(x) x.begin(),x.end()

int N, C[20], V[20];

int main(){
	cin >> N;
	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<N; i++){
		if(C[i]-V[i] > 0) ans += C[i]-V[i];
	}
	cout << ans << endl;
}