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 <iostream>
#include <iomanip>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <algorithm>
#include <cmath>
#include <climits>
#include <math.h>
#include <bitset>

#define X first
#define Y second
#define vsort(v) sort(v.begin(), v.end())
#define vrev(v) reverse(v.begin(), v.end())
#define vmax_p(v) max_element(v.begin(), v.end())
#define vmin_p(v) min_element(v.begin(), v.end())
#define vmax_idx(v) distance(v.begin(), vmax_p(v))
#define vmin_idx(v) distance(v.begin(), vmin_p(v))
#define vcopy(src, dst) copy(src.begin(), src.end(), back_inserter(dst))
#define P pair

// stoi(s): string to int
// stod(s): string to double

// INT_MAX
// INT_MIN
// LLONG_MAX
// LLONG_MIN
// DBL_MIN
// DBL_MAX
// LDBL_MIN
// LDBL_MAX

// A-Z: 65~90
// a-z: 97~122
// |a-z| = 26

// floor(x): truncate

// 2進数変換
// string s = bitset<20>(x).to_string();

using namespace std;
using ll = long long;

int main(int argc, const char *argv[])
{
    int n;
    vector<int> v(n), c(n);
    for (int i = 0; i < n; i++)
    {
        cin >> v[i];
    }
    for (int i = 0; i < n; i++)
    {
        cin >> c[i];
    }

    int sum = 0;
    for (int i = 0; i < n; i++)
    {
        if (v[i] - c[i] > 0)
        {
            sum += v[i] - c[i];
        }
    }
    cout << sum << endl;
}