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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include "stdio.h"
 
int main(void){
	int N, V[50], C[50], res;
	scanf("%d \n", &N);
	if(N != 1){
		for(int i = 0;i < N - 1;i++){
			scanf("%d ", &V[i]);
		}
    }
	scanf("%d \n", &V[N - 1]);
	if(N != 1){
		for(int i = 0;i < N - 1;i++){
			scanf("%d ", &C[i]);
		}
    }
	scanf("%d", &C[N - 1]);
	for(int i = 0;i < N;i++){
		if(V[i] - C[i] > 0){
        	res = res + V[i] - C[i];
        }
	}
	printf("%d", res);
}