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(){
    int n, i;
    int c[51] = {};
    int v[51] = {};
    int ans, max;
    scanf("%d\n", &n);
    for(i=0; i<n; i++){
        scanf("%d", &v[i]);
       /*printf("%d\n", v[i]);*/
    }
    for(i=0; i<n; i++){
        scanf("%d", &c[i]);
    }

    max = 0;
    for(i=0; i<50; i++){
        if(v[i]>c[i]){
            ans = v[i]-c[i];
            max += ans; 
        }
    }
    printf("%d\n", max);
    return 0;
}