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 <iostream>
#include <algorithm>
#include <functional>
#include <vector>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <string>
#include <numeric>
#include <cmath>

#define rep(i, n) for (int i = 0; i < n; i++)
using namespace std;
typedef long long ll;
using P = pair<ll, ll>;

template<class T> inline bool chmin(T& a,T b) {if (a > b) {a = b; return true;} return false;}
template<class T> inline bool chmax(T& a,T b) {if (a < b) {a = b; return true;} return false;}

const ll MOD = 1000000007;
const int INF = 1<<30;
//const ll INF = (ll)1e18 + 1;

int main(){
  int N; cin >> N;
  vector<int> v(N);
  rep(i, N) cin >> v[i];
  vector<int> c(N);
  rep(i, N) cin >> c[i];

  int ans = 0;
  rep(i, N){
    if (v[i] > c[i]) ans += v[i] - c[i];
  }
  cout << ans << endl;
}