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>
#define LENGTH_N 16
#define LENGTH_CV 256
#define NUM_ELEMENT 128

void split_str_space(char str[], char result[][4], int split_num) {
	int i = 0, j = 0, k = 0;
	while (i < LENGTH_CV && j < split_num) {
		if (str[i] == ' ' || str[i] == '\0') {
			while( str[++i] == ' ' || str[i] == '\0'){}
			j++;
			result[j][k] = '\0'; 
			k = 0;
			continue;
		}
		result[j][k++] = str[i++];
	}
}

int main(void) {

	char input_N[LENGTH_N];
	fgets(input_N, LENGTH_N - 1, stdin);
	int N = atoi(input_N);
	
	char input_V[LENGTH_CV];
	char input_C[LENGTH_CV];
	char num_array_V[NUM_ELEMENT][6];
	char num_array_C[NUM_ELEMENT][6];
	
	fgets(input_V, LENGTH_CV - 1, stdin);
	fgets(input_C, LENGTH_CV - 1, stdin);

	split_str_space(input_V, num_array_V, N);
	split_str_space(input_C, num_array_C, N);
	
	int sum = 0;
	for (int i = 0; i < N ; i++){
		int v = atoi(num_array_V[i]);
		int c = atoi(num_array_C[i]);
		if (v - c > 0) sum += v - c;
	}
	
	printf("%d", sum); 

}