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 <algorithm>
#include <iostream>
#include <vector>
#include <queue>
#include <set>
#include <string>
#include <map>
#include <cmath>
#include <cstring>

#define fi first
#define se second
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define rrep(i,n) for (int i = 1; i < (n); ++i)
#define drep(i,n) for (int i = (n)-1; i >= 0; --i)
#define srep(i,s,t) for (int i = s; i < t; ++i)
#define all(x) (x).begin(), (x).end()
#define maxs(x,y) (x = max(x,y))
#define mins(x,y) (x = min(x,y))
#define pb push_back
#define sz(x) (int)(x).size()
#define PQ(T) priority_queue<T, v(T), greater<T> >

using namespace std;
typedef long long ll;
typedef unsigned uint;
typedef unsigned long long ull;
typedef pair<int,int> P;

const ll LINF = 1001002003004005006ll;
const int INF = 1001001001;

const int mod = 1000000007;

int main() {
    int n; cin >> n;
    vector<int> v(n);

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