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>
#define N 20

struct object{
int value;
int cost;
};
struct object a[N];

int totv,maxv=0;
int sel[N]={};

void try(int i);

int main(void)
{ 
    
    int n,i;
 scanf("%d",&n);
    

for (i=0;i < n;i++){ 
    scanf("%d",&a[i].value);}
    
   
for (i=0;i < n;i++){ 
    scanf("%d",&a[i].cost);}
 
    try(n-1);
    
printf("%d\n",maxv);
}

void try(int i)
{ int j;
    sel[i]=0;
 if(i!=0){try(i-1);}
 else{totv=0;
        for(j=0;j<N;j++){if(sel[j]==1)totv+=(a[j].value-a[j].cost);}
        if(totv>maxv)maxv=totv;}
     
 sel[i]=1;
 if(i!=0){try(i-1);}
 else{totv=0;
        for(j=0;j<N;j++){if(sel[j]==1)totv+=(a[j].value-a[j].cost);}
        if(totv>maxv)maxv=totv;}
}