#include<bits/stdc++.h>
using namespace std;
int main(){
  int N;
  cin>>N;
  vector<int64_t>A(N);
  for(int i=0;i<N;i++)
    cin>>A.at(i);
  int64_t ans=0;
  for(int i=1;i<N-1;i++){
  if((N-1)%i==0){
  int k=(N-1)/i; k-=1; k/=2;
    int64_t a=0;
    for(int j=0;j<=k;j++){
    a+=A.at(i*j)+A.at(N-1-i*j);
      if(ans<a)ans=a;
    }
      }else{
  int64_t a=0;
    for(int j=0;(1+j)*i<N-1;j++){
    a+=A.at(i*j)+A.at(N-1-i*j);
      if(ans<a)ans=a;
    }
  }
  }cout<<ans<<endl;
  return 0;
}
### Prompt

Please create a solution in cpp to the following problem:

There is an infinitely large pond, which we consider as a number line. In this pond, there are N lotuses floating at coordinates 0, 1, 2, ..., N-2 and N-1. On the lotus at coordinate i, an integer s_i is written.

You are standing on the lotus at coordinate 0. You will play a game that proceeds as follows:

* 1. Choose positive integers A and B. Your score is initially 0.
* 2. Let x be your current coordinate, and y = x+A. The lotus at coordinate x disappears, and you move to coordinate y.
* If y = N-1, the game ends.
* If y \neq N-1 and there is a lotus floating at coordinate y, your score increases by s_y.
* If y \neq N-1 and there is no lotus floating at coordinate y, you drown. Your score decreases by 10^{100} points, and the game ends.
* 3. Let x be your current coordinate, and y = x-B. The lotus at coordinate x disappears, and you move to coordinate y.
* If y = N-1, the game ends.
* If y \neq N-1 and there is a lotus floating at coordinate y, your score increases by s_y.
* If y \neq N-1 and there is no lotus floating at coordinate y, you drown. Your score decreases by 10^{100} points, and the game ends.
* 4. Go back to step 2.



You want to end the game with as high a score as possible. What is the score obtained by the optimal choice of A and B?

Constraints

* 3 \leq N \leq 10^5
* -10^9 \leq s_i \leq 10^9
* s_0=s_{N-1}=0
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N
s_0 s_1 ...... s_{N-1}


Output

Print the score obtained by the optimal choice of A and B.

Examples

Input

5
0 2 5 1 0


Output

3


Input

6
0 10 -7 -4 -13 0


Output

0


Input

11
0 -4 0 -99 31 14 -15 -39 43 18 0


Output

59

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
int main(){
  int N;
  cin>>N;
  vector<int64_t>A(N);
  for(int i=0;i<N;i++)
    cin>>A.at(i);
  int64_t ans=0;
  for(int i=1;i<N-1;i++){
  if((N-1)%i==0){
  int k=(N-1)/i; k-=1; k/=2;
    int64_t a=0;
    for(int j=0;j<=k;j++){
    a+=A.at(i*j)+A.at(N-1-i*j);
      if(ans<a)ans=a;
    }
      }else{
  int64_t a=0;
    for(int j=0;(1+j)*i<N-1;j++){
    a+=A.at(i*j)+A.at(N-1-i*j);
      if(ans<a)ans=a;
    }
  }
  }cout<<ans<<endl;
  return 0;
}
```