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>
int main(void){
    
    int i,n;
    int v[50];
    int c[100];
    int sum[100];
    int a,b,d;
    
    scanf("%d\n",&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]){
            sum[i]=v[i]-c[i];
        }else{
            sum[i]=0;
        }
    }
    
    a=b=0;
    
    for(i=0; i<n; i++){
        if(sum[i]>a){
            b=a;
            a=sum[i];
        }else if(sum[i]>b){
            b=sum[i];
        }
    }
    d = a+b;
    printf("%d",d);
  return 0;
}