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>
#include <stdlib.h>
#include <string.h>
#include <math.h>

int main(void) {
	int N;
	int C[20];
	int V[20];
	int deff[20];
	int sum;
	int i;
	scanf("%d", &N);
	for(i = 0; i < N; i++){
		scanf("%d ",&C[i]);
	}
	
	for(i = 0; i < N; i++){
		scanf("%d ", &V[i]);
	}
	
	for(i = 0; i < N; i++){
		deff[i] = V[i] - C[i];
		if(deff[i] > 0){
			sum += deff[i];
		}
	}
	
	printf("%d\n", sum);
	
	return 0;
}