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>
#include <stdbool.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <limits.h>
#define MAX 1000                                      
typedef long long ll;
typedef unsigned long long ull;
extern inline int max( int a, int b ) { return a > b ? a : b ; }
extern inline int min( int a, int b ) { return a > b ? b : a ; }
extern inline ll lmax( ll a, ll b ) { return a > b ? a : b ; }
extern inline ll lmin( ll a, ll b ) { return a > b ? b : a ; }
extern inline ull ulmax( ull a, ull b ) { return a > b ? a : b ; }
extern inline ull ulmin( ull a, ull b ) { return a > b ? b : a ; }

int main()
{
	int T, A;
	int ans = 0;
	int arr[20];
	
	scanf("%d", &T);
	for( int i=0; i<T; ++i ){
		scanf("%d", arr+i);
	}
	for( int i=0; i<T; ++i ){
		scanf("%d", &A);
		ans += arr[i] > A ? arr[i]-A : 0;
	}
	printf("%d", ans);
	
}