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>

int main(void)
{
	int i,j,hoge,num,n,x[20],y[20],z[20];
	scanf("%d",&n);
	
	for(i = 0; i < n; i++){
		scanf("%d",&x[i]);
	}
	
	for(i = 0; i < n; i++){
		scanf("%d",&y[i]);
	}
	
	for(i = 0; i < n; i++){
		z[i] = x[i] - y[i];
	}
	
	for(i = 0; i < n-1; i++){
		for(j =i+1 ; j < n; j++){
			if(z[i] < z[j]){
				hoge = z[i];
				z[i] = z[j];
				z[j] = hoge;
			}
		}
	}
	
	num = 0;
	
	for(i = 0; i < n; i++){
		if(z[i] >= 0){
			num = num + z[i];
		}
	}
	
	printf("%d",num);
	
	return 0;
}