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<stdio.h>                                  
                                                   
signed main()                                      
{                                                  
  int i;                                           
  int n;                                           
  scanf("%d",&n);                                  
                                                   
  int v[30];                                       
  for( i = 0; i < n; ++i ){                        
    scanf("%d",&v[i]);                             
  }                                                
                                                   
  int c[30];                                       
  for( i = 0; i < n; ++i ){                        
    scanf("%d",&c[i]);                             
  }                                                
                                                   
  int ans;                                         
  for( i = 0; i < n; ++i ){                        
                                                   
    if( v[i] > c[i] ) ans += ( v[i] - c[i] );      
                                                   
  }                                                
                                                   
  printf("%d\n",ans);                              
                                                   
  return 0;                                        
                                                   
}