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: Wrong Answer)

#include <stdio.h>
#define VMAX 50
#define CMAX 50

int main(void){
	int n=0;	//宝石の個数
	int v[VMAX];
	int c[CMAX];
	int x, y;
	int max = 0;

	//入力
	scanf("%d", &n);
	for(int i = 0; i < n; i++){
		scanf("%d ", v + i);
	}

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

	//計算
	for(int i = 0; i < (n - 1); i++){
		for(int j = i+1; j < n; j++){
			x = v[i] + v[j];
			y = c[i] + c[j];
			if(max < (x - y)) max = (x - y);
		}
	}

	printf("%d", max);

	return 0;

}