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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
 
#define int long long
#define ull unsigned long long
#define ld long double
#define rep(a) rep1(i,a)
#define rep1(i,a) rep2(i,0,a)
#define rep2(i,b,a) for(int i=(b); i<((int)(a)); i++)
#define rep3(i,b,a) for(int i=(b); i>=((int)(a)); i--)
#define all(a) a.begin(),a.end()
#define pii pair<int,int>
#define pb push_back
#define mp make_pair
#define eps 1e-9
#define sz(a) ((int)a.size())
#define pow2(x) (1ll<<(x))
#define ceiling(a,b) (((a)+(b)-1)/(b))
#define print0(a) cout << (a) << ' '
#define print1(a) cout << (a) << '\n'
#define print2(a,b) cout << (a) << ' ',print1(b)
#define print3(a,b,c) cout << (a) << ' ',print2(b,c)
#define print4(a,b,c,d) cout << (a) << ' ',print3(b,c,d)
#ifdef i_am_noob
#define debug(a) print2(#a,a)
#else
#define debug(a) 
#endif
#define ordered_set tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update>
 
using namespace std;
using namespace __gnu_pbds;
 
const int Mod=1000000007,Mod2=998244353;
const int MOD=Mod2;
 
inline int add(int a, int b){return (a+b>=MOD?a+b-MOD:a+b);}
inline int sub(int a, int b){return (a+MOD-b>=MOD?a-b:a+MOD-b);}
int Pow(int a, int b){
	if(!b) return 1;
	int tmp=Pow(a*a%MOD,b>>1);
	return (b&1?tmp*a%MOD:tmp);
}
inline int inverse(int n){return Pow(n,MOD-2);}
 
const int maxn=805;
//i_am_noob
int n,c[20],v[20],ans=0;

signed main(){
	ios_base::sync_with_stdio(0),cin.tie(0);
	cin >> n;
	rep(n) cin >> c[i];
	rep(n) cin >> v[i];
	rep(n) if(c[i]>v[i]) ans+=c[i]-v[i];
	print1(ans);
	return 0;
}