#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long ll;
#define N 100010
ll n,s[N];
ll solve(ll k){
  vector<ll> L,R;
  for(int i=0;i<n;i+=k)L.push_back(s[i]);
  for(int i=n-1;i>=0;i-=k)R.push_back(s[i]);
  ll ans=0,sum=0;
  for(int i=0;i<L.size()&&i<R.size();i++){
    if(((n-1-i*k)%k==0&&(n-1-i*k)<=i*k)||n-1-i*k<=k)break;
    sum+=L[i]+R[i];
    ans=max(ans,sum);
  }
  return ans;
}
int main(){
  cin>>n;
  for(int i=0;i<n;i++)cin>>s[i];
  ll ans=0;
  for(int k=1;k<n;k++){
    ans=max(ans,solve(k));
  }
  cout<<ans<<endl;
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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<iostream>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long ll;
#define N 100010
ll n,s[N];
ll solve(ll k){
  vector<ll> L,R;
  for(int i=0;i<n;i+=k)L.push_back(s[i]);
  for(int i=n-1;i>=0;i-=k)R.push_back(s[i]);
  ll ans=0,sum=0;
  for(int i=0;i<L.size()&&i<R.size();i++){
    if(((n-1-i*k)%k==0&&(n-1-i*k)<=i*k)||n-1-i*k<=k)break;
    sum+=L[i]+R[i];
    ans=max(ans,sum);
  }
  return ans;
}
int main(){
  cin>>n;
  for(int i=0;i<n;i++)cin>>s[i];
  ll ans=0;
  for(int k=1;k<n;k++){
    ans=max(ans,solve(k));
  }
  cout<<ans<<endl;
  return 0;
}

```