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: Accepted)

//     nohomo    //
//  Brankonymous //
//     nohomo    //

#include <bits/stdc++.h>

#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define rep(i,l,r) for (int i=l;i<=r;i++)
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define ll long long
#define mod 1000000007
#define random_jbt mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

using namespace std;

int main(){
    int n,v[30],b[30];
    cin>>n;
    int ans=0;
    for (int i=1;i<=n;i++)cin>>v[i];
    for (int i=1;i<=n;i++){
        cin>>b[i];
        ans+=max(0,v[i]-b[i]);
    }
    cout<<ans;
    return 0;
}