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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#define N 20
int main(void) {
	int v[N] = { 0 };
	int c[N] = { 0 };
	int n = 0;
	int i = 0;
	int sum = 0;


	scanf("%d", &n);

	for (i = 0; i < n; i++) {
		scanf("%d", &v[i]);
	}

	for (i = 0; i < n; i++) {
		scanf("%d", &c[i]);
	}

	for (i = 0; i < n; i++) {
		if ((v[i] - c[i]) >= 0) {
			sum += (v[i] - c[i]);
		}
	}

	printf("%d", sum);
}