Problem:
Score : 700 points

Problem Statement

We have N gemstones labeled 1 through N.

You can perform the following operation any number of times (possibly zero).

Select a positive integer x, and smash all the gems labeled with multiples of x.

Then, for each i, if the gem labeled i remains without getting smashed, you will receive a_i yen (the currency of Japan).
However, a_i may be negative, in which case you will be charged money.

By optimally performing the operation, how much yen can you earn?

Constraints

All input values are integers.

1 \leq N \leq 100

|a_i| \leq 10^9

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

Print the maximum amount of money that can be earned.

Sample Input 1

6
1 2 -6 4 5 3

Sample Output 1

12

It is optimal to smash Gem 3 and 6.

Sample Input 2

6
100 -100 -100 -100 100 -100

Sample Output 2

200

Sample Input 3

5
-1 -2 -3 -4 -5

Sample Output 3

0

It is optimal to smash all the gems.

Sample Input 4

2
-1000 100000

Sample Output 4

99000

Solution in Pascal (FPC 2.6.2) (judge verdict: Wrong Answer)

var  s:array[1..1000]of  boolean;
     a,f:array[1..1000]of  int64;
     n,i,j,x,k,l:longint;
     tot,min:int64;
begin
  readln(n);
  for  i:=1  to  n  do
    begin
      read(a[i]);
      tot:=tot+a[i];
    end;  
  
  fillchar(s,sizeof(s),false);
  for  k:=1  to  n  do
    begin
      min:=99999999999999999;
      fillchar(f,sizeof(f),0);
      for  x:=1  to  n  do
        begin
          if  s[x]=false  then
            for  i:=1  to  n  do
              if  (s[i]=false)and(i mod x=0)  then
                f[x]:=f[x]+a[i];
          if  f[x]<min  then  
            begin
              min:=f[x];        
              l:=x;
            end;  
        end;
      if  min>=0  then  break;    
      for  i:=1  to  n  do
        if  i mod l=0  then
          s[i]:=true;
      tot:=tot-min;      
    end;
  writeln(tot);
end.