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)

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

#define MEM(a, b) memset(a, b, sizeof a);
#define REP(i, k) for(int i = (0); i < (k); i++ )

#define ALL(cont) cont.begin(), cont.end()
#define RALL(cont) cont.begin(), cont.end()
#define FOREACH(it, l) for(auto it=l.begin(); it != l.end(); it++)
#define mp make_pair
#define pb push_back
#define debug puts("FANGO")
#define INF (int)INT_MAX
#define EPS (int)1e-9
#define PI acos(-1)
#define MOD 1000000007
#define endl "\n"

typedef pair<int, int> pii;
typedef pair<int, pii> piii;
typedef vector<int> vi;
typedef vector<string> vs;
typedef vector<pii> vii;
typedef vector<piii> viii;
typedef long long int32;
typedef unsigned long long uint32;
typedef long long int int64;
typedef unsigned long long int uint64;

template <class T>
T sqr(T val) {
  return val * val;
}

void solve(__attribute__((unused)) int tc = 0) {
  int n;
  cin >> n;

  vi v;
  REP(i, n) {
    int x;
    cin >> x;
    v.pb(x);
  }

  vi c;
  REP(i, n) {
    int x;
    cin >> x;
    c.pb(x);
  }

  int ans = 0;
  REP(i, n) {
    if (v[i] - c[i] < 0) continue;
    ans += v[i] - c[i];
  }
  cout << ans <<endl;
  
  return;
}

int main() {
  //freopen("input.txt", "r", stdin);
  //freopen("output.txt", "w", stdout);
  cin.tie(0);
  ios_base::sync_with_stdio(false);

  /*int TC;
  cin >> TC;
  REP(tc, TC) {
    solve(tc);
  }*/
  solve();
}