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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>

int main(void){
    int i;
    int n;
    //printf(">");
    scanf("%d",&n);
    if(n<1||n>20){
        exit(0);
    }
    printf("\n");

    int v[n];
    for(i=0;i<n;i++){
        //printf(">");
        scanf("%d",&v[i]);
        if(v[i]<1||v[i]>50){
            exit(0);
        }
    }
    printf("\n");

    int c[n];
    for(i=0;i<n;i++){
        //printf(">");
        scanf("%d",&c[i]);
        if(c[i]<1||c[i]>50){
            exit(0);
        }
    }
    printf("\n");

    int x=0;
    int y=0;
    x=v[0]+v[n-1];
    y=c[0]+c[n-1];
    int ans=x-y;
    if(ans>=0){
        printf("%d\n",ans);
    /*}else{
        printf("0\n");*/
    }
    return 0;
}