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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;

using ll=long long;
const int inf=100000000;

int in(){
  int a;
  cin >> a;
  return a;
}//int型関数の受け取り

void forin(int* x,int y){
  for (int i=0;i<y;i++){
    cin >> x[i];
  }
}//(x,y) int型配列x[y]の受け取り

int intpow(int x,int y){
  double a=x,b=y;
  return pow(a,b);
}//(x,y) xをy乗

int intfabs(int x){
  double a=x;
  return fabs(a);
}//(x) xの絶対値

int fact(int x){
  if (x == 1) return 1;
  return x*fact(x-1);
}//(x) x!

int fib(int x){
  if (x <= 2) return x;
  int a[x];
  a[0]=1;
  a[1]=2;
  for (int i=2;i<x;i++){
    a[i]=a[i-1]+a[i-2];
  }
  return a[x-1];
}//(x) フィボナッチ数列のx番目

int div1(ll x){
  return x%1000000007;
}

main(){
  int n=in(),v,c,ans=0;
  for (int i=0;i<n;i++){
    cin >> v >> c;
    if (v>c) ans+=(v-c);
  }
  cout << ans << endl;
}